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

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Toolbox:
  • (i)If the rational is improper in nature we can divide and separate the terms to make it a proper rational function.
  • (ii) $\frac{Px+q}{(x+a)(x+b)}=\frac{A}{(x+a)}+\frac{B}{(x+b)}$
  • (iii)$\int \frac{dx}{x+a}=log| x+a |+c$
Given$ I=\int \frac{x^3+x+1}{x^2-1}$
 
Since it is an improper rational function, let us divide
 
$\frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}$
 
$=x+\frac{2x+1}{(x+1)(x-1)}$
 
$\frac{2x+1}{(x+1)(x-1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$
 
2x+1=A(x-1)+B(x+1)
 
Equating the coefficients of x,
 
2=A+B ------(1)
 
Equating the constant terms,
 
1=-A+B -----(2)
 
Now Add equation (1) and equation(2)
 
A+B=2
-A+B=1
_________________
2B=3=>B=3/2
Substitute for B in equ (1)
 
A+B=2
 
$A+\frac{3}{2}=2$
 
A=1/2
 
Hence A=1/2 and B=3/2
 
Now substitude for A and B
 
$\frac{2x+1}{(x+1)(x-1)}=\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$
 
On integrating we get,
 
Therefore $I=\int xdx+\frac{1}{2}\int \frac {dx}{x+1}+\frac{3}{2}\int \frac{dx}{(x-1)}$
 
$\frac{x^2}{2}+\frac{1}{2}log |x+1|+\frac{3}{2} log |x-1|+c$

 

answered Feb 6, 2013 by sreemathi.v
 
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