# Integrate the rational functions$\frac{3x+5}{x^3-x^2-x+1}$

## 1 Answer

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x-a)^2(x-b)}$
• $\;$Form of the partial function$\frac{Ax+B}{(x-a)}+\frac{C}{(x-b)}$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given:$I=\int\frac{3x+5}{x^3-x^2-x+1}dx.$

$\frac{3x+5}{x^3-x^2-x+1}=\frac{3x+5}{(x-1)^2(x+1)}.$

$\qquad\qquad=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}.$

3x+5=A(x-1)(x+1)+B(x+1)+$C(x-1)^2$

$\;\;\;\;\;\;=A(x^2-1)+B(x+1)+C(x^2-2x+1).$

Equating the coefficients of $x^2$

0=A+C-------(1)

Equating the coefficients of x,

3=B-2C-------(2)

Equating the constant terms,

5=-A+B+C------(3)

Multiply equ(1) by 2 and add it with equ(2)
2A+2C=0
B-2C=3
_____________
2A+B=3------(4)
Subtract equ(1) and equ(3)
A+C=0
-A+B+C=5
_________________
2A-B=-5------(5)
Add equ(4) and equ(5)
2A+B=3
2A-B=-5
_____________
4A=-2
A=$\frac{-2}{4}=\frac{-1}{2}.$

Substitute for A in equ(4)

2(-1/2)+B=3.

B=4.

Substitute A in equ (1)

A+C=0.

$\frac{-1}{2}+C=0\Rightarrow c=\frac{1}{2}.$

Hence A=-1/2,B=4,C=1/2.

Hence $\frac{3x+5}{(x-1)^2(x+1)}=\frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}.$

Therefore $I=\frac{-1}{2}\int\frac{1}{(x-1)}dx+4\int\frac{dx}{(x-1)^2}+\frac{1}{2}\int\frac{dx}{(x+1)}$.

On integrating we get,

$\;\;\;=\frac{-1}{2}log|x-1|+4\frac{(x-1)^{-2+1}}+\frac{1}{2}log|x+1|+c.$

$\;\;\;=\frac{-1}{2}log|x+1|-\frac{4}{(x-1)}+\frac{1}{2}log|x-1|+c.$

$log a-log b=log\mid\frac{a}{b}\mid$

$\;\;\;=\frac{1}{2}log\frac{|x+1|}{|x-1|}-\frac{4}{(x-1)}+c.$

answered Feb 6, 2013

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1 answer