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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the integral using substitution\[\int\limits_{-1}^1\frac{dx}{x^2+2x+5}\]

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii)$\int \frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1}(x/a)+c$
Given $\int\limits_{-1}^1\frac{dx}{x^2+2x+5}$
 
Consider $ x^2+2x+5$
 
$ = (x+1)^2+4$
 
$ =(x+1)^2+(2)^2$
 
Hence $\int\limits_{-1}^1\frac{dx}{(x+1)^2+2^2}$
 
This is of the form $\int \frac{dx}{x^2+a^2}=\frac{1}{a} \tan^{-1}(x/a)$
 
Here x=x+1 and a=2 0n integrating;
 
Therefore $\int\limits_{-1}^1\frac{dx}{(x+1)^2+2^2}=\bigg[\frac{1}{2} \tan^{-1}(\frac{x+1}{2})\bigg]_{-1}^1$
 
On applying limits:
 
$=\frac{1}{2}\bigg[\tan^{-1}(\frac{2}{2})-\tan^{-1}(0)\bigg]$
 
$=\frac{1}{2}\tan^{-1}(1)$
 
$=\frac{1}{2} \times \frac{\pi}{4}$
 
$I=\frac{\pi}{8}$

 

answered Feb 12, 2013 by meena.p
 
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