082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

466 Chapter 14 Techniques of integration
4 Find
∫
(a)
∫
(c)
∫
(e)
4x+1
2x 2 +x+3 dx
3
9−2x dx
1
tlnt dt
(b) ∫
(d) ∫
2sin2x
cos2x+7 dx
t
t 2 +1 dt
5 Evaluate the following:
∫ 1
∫
2
π/2
(a)
0 (1 +3x) 2 dx (b) sint √ costdt
0
∫ 2
∫
3 +x 2
(c)
1 x 2 +6x+1 dx (d) e √ t
√ dt
1 t
∫ 2
(e) xsin(π −x 2 )dx
0
Solutions
1 (a)
(4x +1) 8
32
+c (b) − cos(t3 +1)
3
+c
(c) −2e −t2 +c (d) − 3 4 (1 −z)4/3 +c
(e)
1
6 sin6 t+c
2 (a) 3.3588 ×10 5 (b) 0.2 (c) 1.7183
(d) 5.2495 (e) 0.7687
3 103.098
4 (a) ln(2x 2 +x+3)+c
(b) −ln(cos2x +7) +c
(c) − 3 ln(9 −2x) +c
2
1
(d)
2 ln(t2 +1)+c
(e) ln(lnt) +c
5 (a) 0.5 (b) 0.6667 (c) 0.3769
(d) 2.7899 (e) 0.8268
14.4 INTEGRATIONUSINGPARTIALFRACTIONS
The technique of expressing a rational function as the sum of its partial fractions has
been covered in Section 1.7. Some expressions which at first sight look impossible to
integrate mayinfactbeintegrated when expressed astheirpartialfractions.
Example14.14 Find
∫
(a)
∫
(b)
1
x 3 +x dx
13x−4
6x 2 −x−2 dx
Solution (a) Firstexpress the integrand inpartialfractions:
Then,
1
x 3 +x = 1
x(x 2 +1) = A x + Bx+C
x 2 +1
1=A(x 2 +1)+x(Bx+C)
Equating the constant terms: 1 =Aso thatA = 1.
Equating the coefficients ofx: 0 =C so thatC = 0.
Equating the coefficients ofx 2 : 0 =A +B and henceB = −1.