integration

Integral calculus
41
Integration using partial fractions
It was shown in Problem 2, page 19:
41.1 Introduction
2x 2 − 9x − 35
The process of expressing a fraction in terms of (x + 1)(x − 2)(x + 3) ≡ 4
(x + 1) − 3
(x − 2) + 1
(x + 3)
simpler fractions—called partial fractions—is discussed
in Chapter 3, with the forms of partial frac-
2x 2 − 9x − 35
∫
tions used being summarized in Table 3.1, page 18. Hence
(x + 1)(x − 2)(x + 3) dx
Certain functions have to be resolved into partial ∫ {
fractions before they can be integrated as demonstrated
in the following worked problems.
(x + 1) − 3
(x − 2) + 1 }
4
≡
dx
(x + 3)
= 4ln(x + 1) − 3ln(x − 2) + ln(x + 3) + c
{
}
41.2 Worked problems on integration
(x + 1) 4 (x + 3)
or ln
using partial fractions with linear
(x − 2) 3 + c
factors
∫
∫
x
11 − 3x
2 + 1
Problem 1. Determine
x 2 + 2x − 3 dx. Problem 3. Determine
x 2 − 3x + 2 dx.
By dividing out (since the numerator and denominator
are of the same degree) and resolving into partial
As shown in problem 1, page 18:
11 − 3x
x 2 + 2x − 3 ≡ 2
(x − 1) − 5
fractions it was shown in Problem 3, page 19:
(x + 3)
x 2 + 1
x 2 − 3x + 2 ≡ 1 − 2
(x − 1) + 5
(x − 2)
∫
11 − 3x
Hence
x 2 + 2x − 3 dx
∫
x 2 + 1
Hence
∫ {
2
=
(x − 1) − 5 }
x 2 − 3x + 2 dx
∫ {
dx
(x + 3)
≡ 1 − 2
(x − 1) + 5 }
dx
(x − 2)
= 2ln(x − 1) − 5ln(x + 3) + c
= (x − 2) ln(x − 1) + 5ln(x − 2) + c
{ }
(by algebraic substitutions — see Chapter 39)
(x − 2) 5
{ }
or x + ln
(x − 1) 2
(x − 1) 2 + c
or ln
(x + 3) 5 + c by the laws of logarithms
Problem 4. Evaluate
∫
Problem 2. Find
3
x 3 − 2x 2 − 4x − 4
∫
2x 2 2 x
− 9x − 35
2 dx,
+ x − 2
(x + 1)(x − 2)(x + 3) dx correct to 4 significant figures.