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

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x+a)(x^2+b)}\]
  • $\;$Form of the partial function\[\frac{A}{x+a}+\frac{Bx}{x^2+b}\]
  • $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
  • $(iii)\int\frac{dx}{x^2+1}=\tan^{-1}x+c.$
Given $I=\int\frac{x}{(x^2+1)(x-1)}dx.$
 
$\frac{x}{(x^2+1)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}.$
 
$\Rightarrow x=(Ax+B)(x-1)+C(x^2+1).$
 
equating the coefficients of $x^2$,
 
0=A+C.------(1)
 
equating the coefficients of x,
 
1=-A+B------(2)
 
Equating the constant terms,
 
0=-B+C-------(3)
 
Now solving the three equations will give A,B,C eliminate A by adding equ(1) and equ(2)
A+C=0
-A+B=1
_______________
B+C=1--------(4)
Add equ(4) and equ(3)
-B+C=0
B+C=1
_______________
2C=1$\Rightarrow C=\frac{1}{2}$.
 
substitute for C in equ(3),we get
 
$B=\frac{1}{2}.$
 
substitute for B in equ(2),we get
 
$-A+\frac{1}{2}=1\Rightarrow A=\frac{-1}{2}.$
 
$A=\frac{-1}{2},B=\frac{1}{2},C=\frac{1}{2}.$
 
$I=\int\frac{-1/2x+1/2}{(x^2+1)}dx+\int\frac{1/2}{(x-1)}dx.$
 
$\;\;\;=\frac{-1}{2}\int\frac{x}{x^2+1}dx+\frac{1}{2}\int\frac{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$
 
On integrating we get,
 
Let $x^2+1=t.$
 
Therefore $2xdx=dt\Rightarrow xdx=\frac{dt}{2}.$
 
Substituting this we get,
 
$I=\frac{-1}{2}\int\frac{dt/2}{t}+\frac{1}{2}\int{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$
 
$\;\;\;=\frac{-1}{4}\frac{dt}{t}+\frac{1}{2}\int\frac{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$
 
$\;\;\;=\frac{-1}{4}|log t|+\frac{1}{2}\tan^{-1}x+\frac{1}{2}log|x-1|.$
 
Substituting back for t we get,
 
$\;\;\;=\frac{-1}{4}log|x^2+1|+\frac{1}{2}\tan^{-1}x+\frac{1}{2}log|x-1|+c.$

 

answered Feb 5, 2013 by sreemathi.v
edited Feb 5, 2013 by sreemathi.v
 
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