MCQ
$\int_{1/e}^{\tan x} {\frac{{t\,dt}}{{1 + {t^2}}}} + \int_{1/e}^{\cot x} {\frac{{dt}}{{t(1 + {t^2})}}} = $
- A$ - 1$
- ✓$1$
- C$0$
- Dએકપણ નહીં.
$ = \frac{1}{2}\left| {\log (1 + {t^2})} \right|_{1/e}^{\tan x} + \left| {\left\{ {\log t - \frac{1}{2}\log (1 + {t^2})} \right\}} \right|_{1/e}^{\cot x}$
$ = \frac{1}{2}\left[ {\log {{\sec }^2}x - \log \left( {1 + \frac{1}{{{e^2}}}} \right)} \right] + \log \cot x - \log \left( {\frac{1}{e}} \right)$
$ - \frac{1}{2}\left\{ {\log ({\rm{cose}}{{\rm{c}}^2}x) - \log \left( {1 + \frac{1}{{{e^2}}}} \right)} \right\}$
$ = - \log \left( {\frac{1}{e}} \right) = \log e = 1$.
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$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 ?