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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\int\frac{1}{(x^2+1)(x^2+4)}\]

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Toolbox:
  • (i)An rational function is of the form. $\frac{1}{(x^2+a)(x^2+b)}$ can be written as $\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}$
  • (ii)$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
Given $I=\int\frac{1}{(x^2+1)(x^2+4)}dx$
 
Consider $\frac{1}{(x^2+1)(x^2+4)},$ This can be resolved as
 
$\large\frac{(Ax+B)}{(X^2+1)}+\frac{(x+D)}{x^2+4}$
 
$1=(Ax+B)(x^2+4)+(Cx+D)(x^2+1)$
 
$1=Ax^3+4Ax+Bx^2+4B+Cx^3+Cx+Dx^2+D$
 
Equating the coefficient of $x^3$
 
$0=A+C$-----(1)
 
Equating the coefficient of $x^2$
 
$0=B+D$-----(2)
 
Equating the coefficient of $x$
 
$0=4A+C$-----(3)
 
Equating the constant term,
 
$1=4B+D$-----(4)
 
Subtract equ(2) and equ(4)
 
$B+D=0$
$4B+D=1$
___________
$-3B=-1\qquad B=\frac{1}{3}$
 
Substituting for B in equation (2)
 
$B+D=0\qquad Therefore\;\frac{1}{3}+D=0$
 
$=>D=-\frac{1}{3}$
 
Consider equ (1) and equ(3)
 
$A+C=0$
 
$4A+C=0$
 
We can understand that A and C=0
 
Hence $A=0,B=\frac{1}{3},c=0\; and \; d=-\frac{1}{3}$
 
Hence $\frac{1}{(x^2+1)(x^2+4)}=\frac{1}{3(x^2+1)}-\frac{1}{3(x^2+4)}$
 
Therefore $\frac{dx}{(x^2+1)(x^2+4)}=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{1}{x^2+4}dx$
 
This is of the form $\int \frac{dx}{a^2+x^2}=\frac{1}{a} \tan ^{-1}(\frac{x}{a})+c$
 
On integrating we get,
 
$\frac{1}{3}\tan ^{-1}(x)-\frac{1}{3}.\frac{1}{2} \tan ^{-1}\frac{x}{2}+c$
 
$I=\frac{1}{3}\tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\frac{x}{2}+c$

 

 

answered Feb 17, 2013 by meena.p
 
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