Integration by Partial Fraction Decomposition - Louisiana Tech ...

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntegration by Partial FractionDecompositionBernd SchröderBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesIntroductionBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. Every rationalfunction with real coefficientsBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. Every rationalfunction with real coefficients so that the degree of thenumerator is smaller than that of the denominatorBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesTheorem. Partial Fraction Decomposition. 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesProposition.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration constants omitted.)Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration constants omitted.)11. dx = ln|x − c|x − cBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration constants omitted.)11. dx = ln|x − c|∫ x − c12.(x − c) n dx = − 1(n − 1)(x − c) n−1 (n > 1)Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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. (Integration 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.x + 9(x + 3)(x + 4) dx.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.x + 9(x + 3)(x + 4)x + 9(x + 3)(x + 4) dx.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.x + 9=(x + 3)(x + 4)x + 9(x + 3)(x + 4) dx.Ax + 3 +Bx + 4Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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.Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex ZeroesExample.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial fraction decomposition.−36x + 342x 2 + x − 3Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1)Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial fraction decomposition.−36x + 342x 2 + x − 3 = −36x + 34(2x + 3)(x − 1) = A2x + 3 + Bx − 1Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 − 3Partial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.1(x 2 + 1)(x + 3) dx.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.1(x 2 + 1)(x + 3)1(x 2 + 1)(x + 3) dx.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.1(x 2 = Ax + B+ 1)(x + 3) x 2 + 11(x 2 + 1)(x + 3) dx.Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial fraction decomposition.1(x 2 + 1)(x + 3)= Ax + Bx 2 + 1 +1(x 2 + 1)(x + 3) dx.Cx + 3Bernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Compute the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial fraction decomposition.x 3 + 2x 2 + 4x + 10(x 2 + 2x + 2)(x 2 + 4) dxBernd SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science

Background Single Zeroes Higher Degree Numerator Multiple Zeroes Complex Zeroes∫Example. Solve the integralPartial 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech 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 SchröderIntegration by Partial Fraction DecompositionLouisiana Tech University, College of Engineering and Science