Integration by Partial Fraction Decomposition - Louisiana Tech ...
Integration by Partial Fraction Decomposition - Louisiana Tech ...
Integration by Partial Fraction Decomposition - Louisiana Tech ...
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Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong><strong>Decomposition</strong>Bernd SchröderBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroductionBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.2. Rational functions are functions of the formf (x) = a nx n + a n−1 x n−1 + ··· + a 1 x + a 0b m x m + b m−1 x m−1 + ··· + b 1 x + b 0Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.2. Rational functions are functions of the formf (x) = a nx n + a n−1 x n−1 + ··· + a 1 x + a 0b m x m + b m−1 x m−1 + ··· + b 1 x + b 03. They are a natural generalization of polynomialsBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.2. Rational functions are functions of the formf (x) = a nx n + a n−1 x n−1 + ··· + a 1 x + a 0b m x m + b m−1 x m−1 + ··· + b 1 x + b 03. They are a natural generalization of polynomials, and theyoccur pretty frequently in applications, too.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.2. Rational functions are functions of the formf (x) = a nx n + a n−1 x n−1 + ··· + a 1 x + a 0b m x m + b m−1 x m−1 + ··· + b 1 x + b 03. They are a natural generalization of polynomials, and theyoccur pretty frequently in applications, too.4. So we are interested in the integrals of rational functions.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroduction1. Polynomials p(x) = a n x n + a n−1 x n−1 + ··· + a 1 x + a 0 areeasy to integrate.2. Rational functions are functions of the formf (x) = a nx n + a n−1 x n−1 + ··· + a 1 x + a 0b m x m + b m−1 x m−1 + ··· + b 1 x + b 03. They are a natural generalization of polynomials, and theyoccur pretty frequently in applications, too.4. So we are interested in the integrals of rational functions.5. The method we use, partial fraction decomposition, is alsovery important for solving differential equations withLaplace transforms.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Real number version of the FundamentalTheorem of Algebra.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Real number version of the FundamentalTheorem of Algebra. Every polynomial with real coefficientscan be factored into factors of the form (x − c) n and((x − a) 2 + b 2) n .Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficientsBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominatorBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominator can bewritten as a sum of terms of the formBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominator can beAwritten as a sum of terms of the form(x − c) nBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominator can bewritten as a sum of terms of the formAx + B((x − c) 2 + b 2 ) n .A(x − c) n andBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominator can bewritten as a sum of terms of the formA(x − c) n andAx + B((x − c) 2 + b 2 ) n . In each case, n runs from 1 to the exponentthat the respective denominator has in the factorization of thefunction’s denominator.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong>. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominator can bewritten as a sum of terms of the formA(x − c) n andAx + B((x − c) 2 + b 2 ) n . In each case, n runs from 1 to the exponentthat the respective denominator has in the factorization of thefunction’s denominator.So if we can integrate these functions and if we can find a wayto write out the decomposition, then we can integrate rationalfunctions.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther decomposed.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther decomposed. (<strong>Integration</strong> constants omitted.)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|x − cBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan bBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan ∫b1x4.(x 2 + b 2 ) n dx =2(n − 1)b 2 (x 2 + b 2 ) n−1 ++ 2n − 3 ∫12(n − 1)b 2 dx(x 2 + b 2 n−1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan ∫b1x4.(x 2 + b 2 ) n dx =2(n − 1)b 2 (x 2 + b 2 ) n−1 ++ 2n − 3 ∫12(n − 1)b 2 dx (n > 1)(x 2 + b 2 n−1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan ∫b1x4.(x 2 + b 2 ) n dx =2(n − 1)b 2 (x 2 + b 2 ) n−1 ++ 2n − 3 ∫12(n − 1)b 2 dx (n > 1)(x 2 + b 2 n−1∫)x5.x 2 + b 2 dx = 1 (2 ln x 2 + b 2)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan ∫b1x4.(x 2 + b 2 ) n dx =2(n − 1)b 2 (x 2 + b 2 ) n−1 ++ 2n − 3 ∫12(n − 1)b 2 dx (n > 1)(x 2 + b 2 n−1∫)x5.x 2 + b 2 dx = 1 (2 ln x 2 + b 2)∫x6.(x 2 + b 2 ) n dx = − 1 12 (n − 1)(x 2 + b 2 ) n−1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition. Integrals of the rational functions that cannot befurther∫decomposed. (<strong>Integration</strong> constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)∫13.x 2 + b 2 dx = 1 ( x)b arctan ∫b1x4.(x 2 + b 2 ) n dx =2(n − 1)b 2 (x 2 + b 2 ) n−1 ++ 2n − 3 ∫12(n − 1)b 2 dx (n > 1)(x 2 + b 2 n−1∫)x5.x 2 + b 2 dx = 1 (2 ln x 2 + b 2)∫x6.(x 2 + b 2 ) n dx = − 1 12 (n − 1)(x 2 + b 2 ) n−1 (n > 1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.x + 9(x + 3)(x + 4) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.x + 9(x + 3)(x + 4)x + 9(x + 3)(x + 4) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.x + 9=(x + 3)(x + 4)x + 9(x + 3)(x + 4) dx.Ax + 3 +Bx + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.x + 9(x + 3)(x + 4)=x + 9(x + 3)(x + 4)=x + 9(x + 3)(x + 4) dx.Ax + 3 + Bx + 4A(x + 4) + B(x + 3)(x + 3)(x + 4)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)Heaviside ′ s Method :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)x = −3Heaviside ′ s Method :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)x = −3 A = 6Heaviside ′ s Method :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)x = −3 A = 6x = −4Heaviside ′ s Method :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)x = −3 A = 6x = −4Heaviside ′ s Method :B = −5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.<strong>Partial</strong> fraction decomposition.x + 9A=(x + 3)(x + 4) x + 3 + Bx + 4x + 9 A(x + 4) + B(x + 3)=(x + 3)(x + 4) (x + 3)(x + 4)x + 9 = A(x + 4) + B(x + 3)x = −3 A = 6x = −4x + 9(x + 3)(x + 4)=Heaviside ′ s Method :B = −56x + 3 − 5x + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralx + 9(x + 3)(x + 4) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫x + 9(x + 3)(x + 4) dxx + 9(x + 3)(x + 4) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫∫x + 9(x + 3)(x + 4) dx =x + 9(x + 3)(x + 4) dx.6x + 3 − 5x + 4 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫x + 9(x + 3)(x + 4) dx.∫x + 9(x + 3)(x + 4) dx = 6x + 3 − 5x + 4 dx= 6ln|x + 3|Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫x + 9(x + 3)(x + 4) dx.∫x + 9(x + 3)(x + 4) dx = 6x + 3 − 5x + 4 dx= 6ln|x + 3| − 5ln|x + 4|Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫x + 9(x + 3)(x + 4) dx.∫x + 9(x + 3)(x + 4) dx = 6x + 3 − 5x + 4 dx= 6ln|x + 3| − 5ln|x + 4| + cBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral∫x + 9(x + 3)(x + 4) dx.∫x + 9(x + 3)(x + 4) dx = 6x + 3 − 5x + 4 dx= 6ln|x + 3| − 5ln|x + 4| + cBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 32x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 22x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 22x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 16x 4 + 3x 3 − 9x 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 22x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 22x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 22x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x2x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x2x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1− 10x 3 − 5x 2 + 15xBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x2x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15xBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x2x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15xBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x2x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 122x 2 + 11x − 33Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1(−) 22x 2 + 11x − 33Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1(−) 22x 2 + 11x − 33Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1(−) 22x 2 + 11x − 33− 36x + 34Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x22x 2 − 25x + 1(−) 22x 2 + 11x − 33− 36x + 34ThusBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1Thus(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 + x − 322x 2 − 25x + 1(−) 22x 2 + 11x − 33− 36x + 34Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1Thus(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 + x − 322x 2 − 25x + 1(−) 22x 2 + 11x − 33= 3x 2 − 5x + 11− 36x + 34Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 33x 2 − 5x + 112x 2 + x − 3 6x 4 − 7x 3 + 8x 2 − 10x + 1Thus(−) 6x 4 + 3x 3 − 9x 2− 10x 3 + 17x 2 − 10x + 1(−) − 10x 3 − 5x 2 + 15x6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 + x − 322x 2 − 25x + 1(−) 22x 2 + 11x − 33= 3x 2 − 5x + 11 +− 36x + 34−36x + 342x 2 + x − 3 .Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 2 :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 (2 : 88 = A · − 5 )2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 (2 : 88 = A · − 5 ), A = − 17625Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 (2 : 88 = A · − 5 ), A = − 17625−36x + 342x 2 + x − 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 (2 : 88 = A · − 5 ), A = − 17625−36x + 342x 2 + x − 3 = −176 15 2x + 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3<strong>Partial</strong> fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1− 36x + 34 = A(x − 1) + B(2x + 3)x = 1 : − 2 = B · 5, B = − 2 5x = − 3 (2 : 88 = A · − 5 ), A = − 17625−36x + 342x 2 + x − 3 = −176 15 2x + 3 − 2 15 x − 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 + x − 3dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 + x − 3∫= 3x 2 − 5x + 11dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11xBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11x − 176 1ln|2x + 3|5 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11x − 176 15 2 ln|2x + 3| − 2 ln|x − 1|5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11x − 176 15 2 ln|2x + 3| − 2 ln|x − 1| + c5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11x − 176 15 2 ln|2x + 3| − 2 ln|x − 1| + c5= x 3 − 5 2 x2 + 11x − 88 5 ln|2x + 3| − 2 ln|x − 1| + c5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample. Compute the integral∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx.+ x − 3∫ 6x 4 − 7x 3 + 8x 2 − 10x + 12x 2 dx+ x − 3∫= 3x 2 −36x + 34− 5x + 11 +2x 2 + x − 3 dx∫= 3x 2 − 5x + 11 − 176 15 2x + 3 − 2 15 x − 1 dx= x 3 − 5 2 x2 + 11x − 176 15 2 ln|2x + 3| − 2 ln|x − 1| + c5= x 3 − 5 2 x2 + 11x − 88 5 ln|2x + 3| − 2 ln|x − 1| + c5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralx + 3x 2 (x − 1)(x + 2) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2)x + 3x 2 (x − 1)(x + 2) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3x = −2 :1 = D · (−2) 2 · (−3)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3x = −2 : 1 = D · (−2) 2 · (−3), D = − 1 12Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3x = −2 : 1 = D · (−2) 2 · (−3), D = − 1 12x = −1 : 2=A·2+3+ 4 3 + 1 6Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3x = −2 : 1 = D · (−2) 2 · (−3), D = − 1 12x = −1 : 2=A·2+3+ 4 3 + 1 6 , A = −5 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the integralf :=x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) = A x + B x 2 +Cx − 1 +Dx + 2x + 3 = Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx 2 (x + 2) + Dx 2 (x − 1)x = 0 : 3 = B · (−1) · 2, B = − 3 2x = 1 : 4 = C · 1 2 · 3, C = 4 3x = −2 : 1 = D · (−2) 2 · (−3), D = − 1 12x = −1 : 2=A·2+3+ 4 3 + 1 6 , A = −5 (4f = − 5 ) ( 14 x + − 3 )· 12 x 2 + 4 31(−x − 1 + 1 12) 1x + 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integralx + 3x 2 (x − 1)(x + 2) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dxx + 3x 2 (x − 1)(x + 2) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12 x 2 + 4 31(−x − 1 + 1 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12 x 2 + 4 3= − 5 4 ln|x|1(−x − 1 + 1 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12= − 5 4 ln|x| + 3 12 xx 2 + 4 1(−3 x − 1 + 1 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12 x 2 + 4 3= − 5 4 ln|x| + 3 12 x + 4 ln|x − 1|31(−x − 1 + 1 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12 x 2 + 4 1(−3 x − 1 + 1 ) 112 x + 2 dx= − 5 4 ln|x| + 3 12 x + 4 1ln|x − 1| − ln|x + 2|3 12Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12x 2 + 4 1(−3 x − 1 + 1 12= − 5 4 ln|x| + 3 12 x + 4 1ln|x − 1| − ln|x + 2| + c3 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Solve the Integral∫x + 3x 2 (x − 1)(x + 2) dx.x + 3x 2 (x − 1)(x + 2) dx∫ (= − 5 ) ( 14 x + − 3 )· 12x 2 + 4 1(−3 x − 1 + 1 12= − 5 4 ln|x| + 3 12 x + 4 1ln|x − 1| − ln|x + 2| + c3 12) 1x + 2 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)1(x 2 + 1)(x + 3) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 = Ax + B+ 1)(x + 3) x 2 + 11(x 2 + 1)(x + 3) dx.Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 :1 = 10CBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + Cx = −3 : 1 = 10C, C = 110( )x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + Cx = −3 : 1 = 10C, C = 110x = 0 :( )x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + Cx = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110( )x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 310Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 310x = 1 :Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 3x = 1 : 1 =(A + 3 10) 104 + 1 10 · 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 3x = 1 : 1 =(A + 3 10) 104 + 1 10 · 2, A = − 1 10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 3x = 1 : 1 =(A + 3 10) 104 + 1 10 · 2, A = − 1 101(x 2 + 1)(x + 3)=Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 3x = 1 : 1 =(A + 3 10) 104 + 1 10 · 2, A = − 1 101(x 2 + 1)(x + 3)= 1 −x + 310 x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral<strong>Partial</strong> fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 31 = (Ax + B)(x + 3) + C( )x 2 + 1x = −3 : 1 = 10C, C = 110x = 0 : 1 = 3B + 110 , B = 3x = 1 : 1 =(A + 3 10) 104 + 1 10 · 2, A = − 1 101(x 2 + 1)(x + 3)= 1 −x + 310 x 2 + 1 + 1101x + 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1 + 310 arctan(x)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1 + 310 arctan(x) + 1 ln|x + 3|10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1 + 310 arctan(x) + 1 ln|x + 3| + c10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1 + 310 arctan(x) + 1 ln|x + 3| + c10)= − 120 ln (x 2 + 1+ 310 arctan(x) + 1 ln|x + 3| + c10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integral1(x 2 + 1)(x + 3) dx.∫1(x 2 + 1)(x + 3) dx∫ 1 −x + 3=10 x 2 + 1 + 1 110 x + 3 dx∫= − 1 x10 x 2 + 1 + 3 110 x 2 + 1 + 1 110 x + 3 dx= − 1 1( )10 2 ln x 2 + 1 + 310 arctan(x) + 1 ln|x + 3| + c10)= − 120 ln (x 2 + 1+ 310 arctan(x) + 1 ln|x + 3| + c10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.Y := x3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4)=x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.Y := x3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4)=x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxAx + Bx 2 + 2x + 2 + Cx + Dx 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.Y := x3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4)=x 3 + 2x 2 + 4x + 10 =x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxAx + Bx 2 + 2x + 2 + Cx + Dx 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2+(4A + 2C + 2D)xBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2+(4A + 2C + 2D)x + (4B + 2D)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2A + C = 1+(4A + 2C + 2D)x + (4B + 2D)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2A + C = 1B + 2C + D = 2+(4A + 2C + 2D)x + (4B + 2D)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2A + C = 1B + 2C + D = 24A + 2C + 2D = 4+(4A + 2C + 2D)x + (4B + 2D)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxY := x3 + 2x 2 + 4x + 10 Ax + B(x 2 + 2x + 2)(x 2 =+ 4) x 2 + 2x + 2 + Cx + Dx 2 + 4( ) ( )x 3 + 2x 2 + 4x + 10 = (Ax+B) x 2 +4 + (Cx+D) x 2 +2x+2x 3 + 2x 2 + 4x + 10 = (A + C)x 3 + (B + 2C + D)x 2A + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10+(4A + 2C + 2D)x + (4B + 2D)Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.A + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2D = − 1 5 ,Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2D = − 1 5 , C = −1 5 ,Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2D = − 1 5 , C = −1 5 , B = 13 5 ,Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2D = − 1 5 , C = −1 5 , B = 13 5 , A = 6 5 ,Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral<strong>Partial</strong> fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxA + C = 1B + 2C + D = 24A + 2C + 2D = 44B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 8C − 2D = 2A + C = 1B + 2C + D = 2− 2C + 2D = 04B + 2D = 10A + C = 1B + 2C + D = 2− 2C + 2D = 0− 10D = 2D = − 1 5 , C = −1 5 , B = 13 5 , A = 6 65 , Y = 5x + 135x 2 + 2x + 2 + − 1 5 x − 1 5x 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx=∫ 65x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx==∫ 65x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 15 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫ 6= 5x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 1=5 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dx= 6 1( )5 2 ln (x + 1) 2 + 1Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫ 6= 5x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 1=5 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dx= 6 1( )5 2 ln (x + 1) 2 + 1 + 7 arctan(x + 1)5Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫ 6= 5x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 1=5 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dx= 6 1( )5 2 ln (x + 1) 2 + 1 + 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx==∫ 65x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 15 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dx( )2 ln (x + 1) 2 + 1 + 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 4− 1 1( x)5 2 arctan 2= 6 15Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx==∫ 65x + 13 5x 2 + 2x + 2 + − 5 1x − 1 5x 2 + 4dx∫ 6 x + 15 (x + 1) 2 + 1 + 7 15 (x + 1) 2 + 1 − 1 x5 x 2 + 4 − 1 15 x 2 + 4 dx( )2 ln (x + 1) 2 + 1 + 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 4− 1 1( x)5 2 arctan + c2= 6 15Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralx 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx= 6 1( )5 2 ln (x + 1) 2 + 1 + 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 4− 1 1( x)5 2 arctan + c2Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx= 6 1( )5 2 ln (x + 1) 2 + 1− 1 1( x)5 2 arctan + c2= 3 )((x5 ln + 1) 2 + 1 + 7 5− 1 ( x)10 arctan + c2x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx+ 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 41( )arctan(x + 1) −10 ln x 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science
Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integral∫x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx= 6 1( )5 2 ln (x + 1) 2 + 1− 1 1( x)5 2 arctan + c2= 3 )((x5 ln + 1) 2 + 1 + 7 5− 1 ( x)10 arctan + c2x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dx+ 7 5 arctan(x + 1) − 1 1( )5 2 ln x 2 + 41( )arctan(x + 1) −10 ln x 2 + 4Bernd Schröder<strong>Integration</strong> <strong>by</strong> <strong>Partial</strong> <strong>Fraction</strong> <strong>Decomposition</strong><strong>Louisiana</strong> <strong>Tech</strong> University, College of Engineering and Science