$ = \int_0^{\pi /4} {\frac{{\cos x\,dx}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} $
$ = \frac{1}{3}\int_0^{1/\sqrt 2 } {\left( {\frac{1}{{1 - {t^2}}} + \frac{2}{{1 + 2{t^2}}}} \right)} \,dt$
By partial fractions, where $t = \sin x$
$ = \frac{1}{3}\left[ {\frac{1}{{2.1}}\log \frac{{1 + t}}{{1 - t}} + \frac{2}{{\sqrt 2 }}{{\tan }^{ - 1}}t\sqrt 2 } \right]_0^{1/\sqrt 2 }$
$ = \frac{1}{3}\left[ {\frac{1}{2}\log \frac{{(\sqrt 2 + 1)}}{{(\sqrt 2 - 1)}} + \sqrt 2 {{\tan }^{ - 1}}1} \right]$
$ = \frac{1}{3}\left[ {\frac{1}{2}\log {{(\sqrt 2 + 1)}^2} + \sqrt 2 .\frac{\pi }{4}} \right]$
$ = \frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$.
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$f(x)=3 \log _{e}\left|\frac{x-1}{x+1}\right|-\frac{2}{x-1}$
મુજબ આપેલ છે. તો નીચેનામાંથી કયા અંતરાલોમાં વિધેય $f ( x )$ વધે છે ?