integration

410 INTEGRAL CALCULUS
∫ ∫ {
}
2x + 3
Thus
(x − 2) 2 dx ≡ 2
(x − 2) + 7
(x − 2) 2 dx
= 2ln(x − 2) − 7
(x − 2) + c
⎡∫
⎤
7
dx is determined using the algebraic
⎣ (x − 2) 2 ⎦
substitution u = (x − 2) — see Chapter 39.
Problem 6.
Find
∫ 5x 2 − 2x − 19
(x + 3)(x − 1) 2 dx.
It was shown in Problem 6, page 21:
5x 2 − 2x − 19
(x + 3)(x − 1) 2 ≡ 2
(x + 3) + 3
(x − 1) − 4
(x − 1) 2
∫ 5x 2 − 2x − 19
Hence
(x + 3)(x − 1) 2 dx
∫ {
2
≡
(x + 3) + 3
}
(x − 1) − 4
(x − 1) 2 dx
= 2ln(x + 3) + 3ln(x − 1) + 4
(x − 1) + c
{
or ln (x + 3) 2 (x − 1) 3} + 4
(x − 1) + c
Problem 7.
∫ 1
−2
Evaluate
3x 2 + 16x + 15
(x + 3) 3 dx,
correct to 4 significant figures.
It was shown in Problem 7, page 22:
3x 2 + 16x + 15
(x + 3) 3 ≡
−2
3
(x + 3) − 2
(x + 3) 2 − 6
(x + 3) 3
∫ 3x 2 + 16x + 15
Hence
(x + 3) 3 dx
∫ 1
{
}
3
≡
(x + 3) − 2
(x + 3) 2 − 6
(x + 3) 3 dx
=
=
[
3ln(x + 3) + 2
(
3ln4+ 2 4 + 3 16
]
(x + 3) + 3 1
(x + 3) 2 −2
) (
− 3ln1+ 2 1 + 3 )
1
= −0.1536, correct to 4 significant figures
Now try the following exercise.
Exercise 164 Further problems on integration
using partial fractions with repeated
linear factors
In Problems 1 and 2, integrate with respect
to x.
∫ 4x − 3
1.
(x + 1) 2 dx [
4ln(x + 1) + 7 ]
(x + 1) + c
∫ 5x 2 − 30x + 44
2.
(x − 2) 3 dx
⎡
5ln(x − 2) + 10 ⎤
⎢
(x − 2) ⎥
⎣
2 ⎦
−
(x − 2) 2 + c
In Problems 3 and 4, evaluate the definite integrals
correct to 4 significant figures.
3.
∫ 2
1
∫ 7
x 2 + 7x + 3
x 2 (x + 3)
0
[1.663]
18 + 21x − x 2
4.
dx [1.089]
6 (x − 5)(x + 2) 2
∫ 1
( 4t 2 )
+ 9t + 8
5. Show that
(t + 2)(t + 1) 2 dt = 2.546,
correct to 4 significant figures.
41.4 Worked problems on integration
using partial fractions with
quadratic factors
Problem 8.
∫ 3 + 6x + 4x 2 − 2x 3
Find
x 2 (x 2 dx.
+ 3)