_ ,0 ^ , xdx (1 + x)(1 + x^2 ) = - Vidyadip

MCQ

$\int_{\,0}^{\,\infty } {\frac{{xdx}}{{(1 + x)(1 + {x^2})}} = } $

A
$0$
B
$\pi /2$
✓
$\pi /4$
D
$1$

✓

Answer

Correct option: C.

$\pi /4$

(c) $\int_0^\infty {\frac{{xdx}}{{(1 + x)(1 + {x^2})}} = \int_0^\infty {\frac{{ - \frac{1}{2}dx}}{{(1 + x)}} + \int_0^\infty {\frac{{\left( {\frac{1}{2}x + \frac{1}{2}} \right)}}{{1 + {x^2}}}dx} } } $

$ = \left[ {\frac{{ - 1}}{2}\log (1 + x)} \right]_0^\infty + \frac{1}{2} \times \frac{1}{2}[\log \,(1 + {x^2})]\,_0^\infty + \frac{1}{2}[{\tan ^{ - 1}}x]\,_0^\infty $

$ = 0 + 0 + \frac{1}{2}\left[ {\frac{\pi }{2} - 0} \right] = \frac{\pi }{4}$.

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