STD 12 - 7.1 inderfinite integral — JEE कक्षा 11 साइन्स — Question

$\int_{}^{} {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{{{\left( {x + \frac{1}{x}} \right)}^2} - 1}}} \,dx$
 $x + \frac{1}{x} = t  $ रखने पर $ \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)\,dx = dt,$ 
$\int_{}^{} {\frac{{dt}}{{{t^2} - 1}}} = \frac{1}{2}\log \left| {\frac{{t - 1}}{{t + 1}}} \right| + c$
$ = \frac{1}{2}\log \left( {\frac{{x + \frac{1}{x} - 1}}{{x + \frac{1}{x} + 1}}} \right) + c = \frac{1}{2}\log \left( {\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}}} \right) + c.$