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

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)^2(x-b)}\]
  • $\;$Form of the partial function\[\frac{A}{(x-a)}+\frac{B}{(x-a)^2}+\frac{C}{(x-b)}\]
  • $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given $I=\int\frac{x}{(x-1)^2(x+2)}dx.$
 
$\frac{x}{(x-1)^2(x+2)}=\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}.$
 
$\Rightarrow x=A(x-1)(x+2)+B(x+2)+C(x-1)^2.$
 
$x=A(x^2+x-2)+B(x+2)+C(x^2-2x+1).$
 
Equating the coefficients of $x^2$,
 
A+C=0-----(1)
 
Equating the coefficients of x,
 
1=A+B-2C-----(2)
 
Equating the constant terms,
 
0=-2A+2B+C-------(3)
 
On solving (1),(2) and (3) we obtain A,B, and C
 
Subtract equ(1) and equ(2)
A+C=0
A+B-2C=1
_________________
-B+3C=-1-------(4)
Multiply equ(1) by equ(2) and add it with equ(3),
 
2A+2C=0
2A+2B+C=0
___________________
2B+3C=0-------(5)
Multiply equ(4) by (2) and add it with equ(5)
 
2B+3C=0
-2B+6C=-2
________________
9C=-2
$\Rightarrow C=\frac{-2}{9}.$
 
Substituting for C in equ(1) we get,
 
A+C=0.
 
$A-\frac{1}{9}=0\Rightarrow A=\frac{2}{9}.$
 
Substitute for C in equ(5) we get,
 
2B+3(-2/9)=0.
 
2B=6/9$\Rightarrow B=\frac{1}{3}.$
 
Hence A=2/9,B=1/3,C=-2/9.
 
Now substituting for A,B and C we get,
 
$I=\int\begin{bmatrix}\frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}\end{bmatrix}dx.$
 
$\;\;\;=\frac{2}{9}\int\frac{dx}{(x-1)}+\frac{1}{3}\int\frac{dx}{(x-1)^2}-\frac{2}{9}\int\frac{dx}{(x+2}.$
 
On integrating we get,
 
$\;\;\;=\frac{2}{9}log|x+1|+\frac{1}{3}\big(\frac{(x-1)^-2+1}{-2+1}\big)-\frac{2}{9}log|x+2|+c.$
 
$\;\;\;=\frac{2}{9}log\frac{|x-1|}{|x+2|}|-\frac{1}{3(x-1)}+c.$

 

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