MCQ
$\int_0^\infty {\frac{{\log \,(1 + {x^2})}}{{1 + {x^2}}}} \,dx = $
- A$\pi \log \frac{1}{2}$
- ✓$\pi \log 2$
- C$2\pi \log \frac{1}{2}$
- D$2\pi \log 2$
Put $x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta ,$
$\therefore $ $I = \int_0^{\pi /2} {\log {{(\sec \theta )}^2}d\theta = 2\int_0^{\pi /2} {\log \sec \theta \,\,d\theta } } $
$ = - 2\int_0^{\pi /2} {\log \cos \theta \,\,d\theta = - 2.\,\,\frac{\pi }{2}\log \frac{1}{2}} $
$ = - \pi \log \frac{1}{2} = \pi \log 2$.
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