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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integral\[\int\limits_2^3\frac{dx}{x^2-1}\]

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Toolbox:
  • $\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int \large \frac{dx}{x^2-a^2}=\frac{1}{2a}log \bigg|\frac{x-a}{x+a}\bigg|+c$
Given $ I=\int \limits_2^3 \large \frac{dx}{x^2-1}$
 
On integrating we get,Here a=1
 
$\large\frac{1}{2 \times 1} \bigg[log \bigg|\frac{x-a}{x+a}\bigg|\bigg]^3_2$
 
$=\bigg[ \large \frac{1}{2}log \bigg|\frac{x-a}{x+a}\bigg|\bigg]^3_2$
 
On Applying limits we get,
 
$\large\frac{1}{2} log \bigg[ \bigg|\frac{3-1}{3+1}\bigg|-\bigg|\frac{2-1}{2+1}\bigg|\bigg]$
 
$I=\frac{1}{2} log \bigg[\frac{2}{4}-\frac{1}{3}\bigg]$
 
But we know $log a-log b=log (a/b)$
 
$ I=\large \frac{1}{2} log (\frac{1}{2} \div \frac{1}{3})=\frac{1}{2} log (\frac{1}{2} \times 3)$
 
$\int \limits^2_3 \large\frac{dx}{x^2-1}=\frac {1}{2} log (3/2)$

 

answered Feb 11, 2013 by meena.p
 
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