_0^ ,(1 + x^2 ) 1 + x^2 ,dx = - Vidyadip

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
$2\pi \log 2$

✓

Answer

Correct option: B.

$\pi \log 2$

b
(b) Let $I = \int_0^\infty {\frac{{\log (1 + {x^2})}}{{1 + {x^2}}}\,\,dx} $

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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