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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{3x+5}{x^3-x^2-x+1}\]

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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 by sreemathi.v
 
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