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

14.4 Integration using partial fractions 467
(b)
Then, ∫ ∫
1 1
x 3 +x dx = x − x
x 2 +1 dx
=ln|x|− 1 2 ln|x2 +1|+c=ln
x
∣√ x2 +1
∣ +c
∫
∫
13x−4
6x 2 −x−2 dx = 13x−4
(2x +1)(3x −2) dx
∫
=
= 3 2
3
2x+1 + 2
3x−2 dx
∫
2
2x+1 dx + 2 ∫
3
using partialfractions
3
3x−2 dx
= 3 2 ln|2x+1|+2 3 ln|3x−2|+c
∫ 1
4t
Example14.15 3 −2t 2 +3t−1
Evaluate dt.
0 2t 2 +1
Solution Usingpartialfractions we maywrite
4t 3 −2t 2 +3t−1 t
=2t−1+
2t 2 +1 2t 2 +1
Hence,
∫ 1
0
4t 3 −2t 2 +3t−1
2t 2 +1
dt =
=
∫ 1
0
2t−1+
[
t
2t 2 +1 dt = t 2 −t + 1 ] 1
4 ln |2t2 +1|
0
[1 −1+ 1 ]
4 ln3 −
[0 −0+ 1 ]
4 ln1 = 0.275
EXERCISES14.4
1 Bywriting the integrand asits partialfractions find
∫ x +3
(a)
x 2 +x dx
∫ t −3
(b)
t 2 −1 dt
∫
8x+10
(c)
4x 2 +8x+3 dx
∫ 2t 2 +3t+3
(d) dt
2(t +1)
∫ 2x 2 +x+1
(e)
x 3 +x 2 dx
2 Evaluate the following integrals:
∫ 3 5x+6
(a)
1 2x 2 +4x dx
∫ 1 3x+5
(b)
0 (x +1)(x +2) dx
∫ 2 3−3x
(c)
1 2x 2 +6x dx
∫ 0 4x+1
(d)
−1 2x 2 +x−6 dx
∫ 3 x 2 +2x−1
(e)
2 (x 2 +1)(x −1) dx