INTEGRATION USING PARTIAL FRACTIONS 409H
By dividing out and resolving into partial fractions
it was shown in Problem 4, page 20:
x^3 − 2 x^2 − 4 x− 4
x^2 +x− 2≡x− 3 +4
(x+2)−3
(x−1)Hence
∫ 32x^3 − 2 x^2 − 4 x− 4
x^2 +x− 2dx≡∫ 32{
x− 3 +4
(x+2)−3
(x−1)}
dx=[
x^2
2− 3 x+4ln(x+2)−3ln(x−1)] 32=(
9
2− 9 +4ln5−3ln2)−(2− 6 +4ln4−3ln1)=− 1. 687 , correct to 4 significant figuresNow try the following exercise.
Exercise 163 Further problems on integra-
tion using partial fractions with linear factorsIn Problems 1 to 5, integrate with respect tox1.∫
12
(x^2 −9)dx⎡⎢
⎣2ln(x−3)−2ln(x+3)+cor ln{
x− 3
x+ 3} 2
+c⎤⎥
⎦2.∫
4(x−4)
(x^2 − 2 x−3)dx⎡⎢
⎣5ln(x+1)−ln (x−3)+cor ln{
(x+1)^5
(x−3)}+c⎤⎥
⎦3.∫
3(2x^2 − 8 x−1)
(x+4)(x+1)(2x−1)dx⎡⎢
⎢
⎢
⎣7ln(x+4)−3ln(x+1)
−ln (2x−1)+c orln{
(x+4)^7
(x+1)^3 (2x−1)}
+c⎤⎥
⎥
⎥
⎦4.∫
x^2 + 9 x+ 8
x^2 +x− 6dx[
x+2ln(x+3)+6ln(x−2)+c
orx+ln{(x+3)^2 (x−2)^6 }+c]5.∫
3 x^3 − 2 x^2 − 16 x+ 20
(x−2)(x+2)dx⎡⎣3 x^2
2− 2 x+ln (x−2)−5ln(x+2)+c⎤⎦In Problems 6 and 7, evaluate the definite inte-
grals correct to 4 significant figures.6.∫ 43x^2 − 3 x+ 6
x(x−2)(x−1)dx [0.6275]7.∫ 64x^2 −x− 14
x^2 − 2 x− 3dx [0.8122]- Determine the value of k, given that:
∫ 1
0(x−k)
(3x+1)(x+1)dx= 0[
1
3]- The velocity constantkof a given chemical
reaction is given by:
kt=∫ (
1
(3− 0. 4 x)(2− 0. 6 x))
dxwherex=0 when t = 0. Show that:kt=ln{
2(3− 0. 4 x)
3(2− 0. 6 x)}41.3 Worked problems on integration
using partial fractions with
repeated linear factorsProblem 5. Determine∫
2 x+ 3
(x−2)^2dx.It was shown in Problem 5, page 21:2 x+ 3
(x−2)^2≡2
(x−2)+7
(x−2)^2