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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{1}{x^4-1}\]

$\begin{array}{1 1} \frac{1}{4} \log \big | \frac {x-1}{x+1} \big | -\frac{1}{2}\tan^{-1}x+c \\ \frac{1}{4} \log \big | \frac {x+1}{x-1} \big | -\frac{1}{4}\tan^{-1}x+c \\\frac{1}{4} \log \big | \frac {x^2}{x-1} \big | -\frac{1}{2}\tan^{-1}x+c \\ \frac{1}{4} \log \big | \frac {x^2}{x+1} \big | +\frac{1}{2}\tan^{-1}x+c\end{array} $

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1 Answer

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x+a)(x^2+b)}\]
  • $\;$Form of the partial function\[\frac{A}{(x+a)}+\frac{Bx+C}{(x^2+b)}\]
  • $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
  • $(iii)\;\int\frac{dx}{(x^2+b}=\frac{1}{b}\tan^{-1}\big(\frac{x}{b}\big)+c.$
Given:$I=\int\frac{1}{x^4-1}dx.$
 
$\frac{1}{x^4-1}=\frac{1}{(x^2-1)(x^2+1)}=\frac{1}{(x+1)(x-1)(x^2+1)}.$
 
Therefore $\frac{1}{(x+1)(x-1)(x^2+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{(x+1)}{(x^2+1)}.$
 
$\Rightarrow 1=A(x-1)(x^2+1)+B(x+1)(x^2+1)+(x+D)(x+1)(x-1).$
 
$\;\;\;=A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+(x+D)(x^2-1).$
 
Equating the coefficients of $x^3$,
 
0=A+B+C------(1)
 
Equating the coefficients of $x^2$,
 
0=-A+B+D-----(2)
 
Equating the coefficients of x,
 
0=A+B-C-------(3)
 
Equating the constant terms,
 
1=-A+B-D-----(4)
 
Let us take equ(1) and equ(2) and eliminate A by adding them.
A+B+C=0
-A+B+D=0
________________
2B+C+D=0------(5)
Let us take equ(2) and equ(3) and eliminate A by adding then,
-A+B+D=0
A+B-C=0
_______________
2B+D-C=0------(6)
Now add (5) and (6) to eliminate C
2B+C+D=0
2B-C+D=0
__________________
4B+2D=0
$\Rightarrow$ 2B+D=0------(7)
 
Now subtract equ(1) and equ(4) to elliminate A and B
 
-A+B+D=0
 
-A+B-D=1
 
2D=-1$\Rightarrow D=\frac{-1}{2}.$
 
Substitute for D in equ(7)
 
2B+D=0
 
2B-$\frac{1}{2}$=0.
 
2B=$\frac{1}{2}\Rightarrow B=1/4.$
 
Substitute for B and D in equ(6)
 
2(1/4)+(-1/2)-C=0$\Rightarrow C=0.$
 
Now substitute for B and C in equ(1)
 
A+B+C=0.
 
A+(1/4)+0=0.
 
A=-1/4.
 
Hence A=-1/4,B=1/4,C=0 and D=-1/2.
 
Now substitute for A,B,C and D in I
 
$\frac{1}{(x-1)(x+1)(x^2+1)}=-\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+\frac{0-1/2}{(x^2+1)}$.
 
$I=\frac{-1}{4}\int\frac{dx}{(x+1)}+\frac{1}{4}\int\frac{dx}{(x-1)}-\frac{-1}{2}\int\frac{dx}{(x^2+1)}.$
 
On integrating we get,
 
$\;\;\;=\frac{-1}{4}log|x+1|+\frac{1}{4}log|x-1|-\frac{1}{2}\tan^{-1}x+c.$
 
log a-log b=log$\mid\frac{a}{b}\mid$,similarly
 
$\;\;\;=\frac{1}{4}log\frac{|x-1|}{|x+1|}-\frac{1}{2}\tan^{-1}x+C.$
 

 

answered Feb 6, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans
 
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