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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{x}{(x-1)(x-2)(x-3)}\]

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)(x-b)(x-c)},a\neq b\neq c\]
  • $(ii)\;$Form of the partial function\[\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-c)}\]
  • $(iii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given $I=\int\frac{x}{(x-1)(x-2)(x-3)}dx.$
 
This can be written as:
 
$\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}.$
 
$\Rightarrow x=A(x-2)(x-3)+B(x-1)(x-3)+c(x-1)(x-2)$.
 
Equating the coefficients of $x^2$,
 
0=A+B+C-----(1)
 
equating the coefficients of x,
 
1=-5A-4B-3C-------(2)
 
Equating the constant terms,
 
0=6A+3B+2C------(3)
 
Multiply equ(1) by 5 and subtract equ(1) after multiplying by 5 from (2)
 
we get B+2C=1-------(4)
 
similarly multiply equ(1) by 6 and subtract it from (3) we get,
 
3B+4C=0-----(5)
 
Multiply equ(4) by 3 and subtract from equ(5)
 
we get $C=\frac{3}{2}$ and B=-2.
 
Substitute for B and C in equation (1)
 
We get,
 
A-2+$\frac{3}{2}$=0.
 
$\Rightarrow A=2-\frac{3}{2}=\frac{1}{2}.$
 
Hence $A=\frac{1}{2};B=-2$ and $C=\frac{3}{2}.$
 
Hence $I=\frac{1}{2}\int\frac{dx}{x-1}-2\int\frac{dx}{x-2}+\frac{3}{2}\int\frac{dx}{x-3}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}log|x-1|-2log|x-2|+\frac{3}{2}log|x-3|+c.$

 

answered Feb 5, 2013 by sreemathi.v
 
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