MCQ
$\int_{\,0}^{\,\infty } {\,\log \left( {x + \frac{1}{x}} \right)\frac{{dx}}{{1 + {x^2}}}} $ is equal to
- ✓$\pi \log 2$
- B$ - \pi \log 2$
- C$(\pi /2)\log 2$
- D$ - (\pi /2)\log 2$
Put $x = \tan \theta \Rightarrow \,\,dx = {\sec ^2}\theta \,\,d\theta $
$ \Rightarrow I = \int_0^{\pi /2} {\,\,\,\,\,\log (\tan \theta + \cot \theta } )\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta }}\,d\theta $
==> $I = \int_0^{\pi /2} {\,\,\,\,\,\log (\tan \theta + \cot \theta } )d\theta $
$ \Rightarrow I = \int_0^{\pi /2} {\log \frac{{(1 + {{\tan }^2}\theta )}}{{\tan \theta }}\,d\theta } $
==> $I$ $ = 2\int_0^{\pi /2} {\log \sec \theta \,d\theta - \int_0^{\pi /2} {\log \tan \theta } } \,d\theta $
==> $I$ $ = 2\int_0^{\pi /2} {\log \sec \theta \,\,d\,\theta } $;
$\left\{ \,\because \int_{0}^{\pi /2}{\log \tan \theta =0} \right\}$
$ \Rightarrow \,I = - 2\int_0^{\pi /2} {\,\,\,\,\,\log \cos \theta \,d\theta } $
==>$I = - 2 \times \frac{{ - \pi }}{2}\log 2$,$\left\{ \because \int_{0}^{\pi /2}{\log \cos \theta =-\frac{\pi }{2}\log 2} \right\}$
==> $I = \pi \log 2$.
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