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

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)(x-b)},a\neq b\]
  • $(ii)\;$Form of the partial function\[\frac{A}{x-a}+\frac{B}{x-b}\]
  • $(iii)\;\int\frac{dx}{x-a}=log(x-a)+c.$
Given $I=\int\frac{x}{(x+1)(x+2)}dx.$
 
This can be written as,\[\frac{x}{(x+1)(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}\]
 
(ie)x=A(x+2)+B(x+1).
 
Equating the coefficients of x,
 
1=A+B-------(1)
 
Equating the constant terms
 
0=2A+B------(2)
 
Now solving (1) and (2)
 
A+B=1------(1)
2A+B=0-----(2)
_______________
On subtracting (1) and (2) we get,
 
-A=1.Therefore A=-1.
 
Substituting for A in (1)
 
We get B=2.
 
We get A=-1 and B=2.
 
Therefore $I=\int\frac{-1}{(x+1)}dx+2\int\frac{dx}{(x+2)}.$
 
On integrating we get,
 
$\;\;\;=-log|x+1|+2log|x+2|+c.$
 
$\;\;\;=log\frac{(x+2)^2}{|x+1|}+c.$

 

answered Feb 5, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans
 
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