A) 0
B) \[\pi /2\]
C) \[\pi /4\]
D) 1
Correct Answer: C
Solution :
\[\int_{0}^{\infty }{\frac{xdx}{(1+x)(1+{{x}^{2}})}=\int_{0}^{\infty }{\frac{-\frac{1}{2}dx}{(1+x)}+\int_{0}^{\infty }{\frac{\left( \frac{1}{2}x+\frac{1}{2} \right)}{1+{{x}^{2}}}dx}}}\] \[=\left[ \frac{-1}{2}\log (1+x) \right]_{0}^{\infty }+\frac{1}{2}\times \frac{1}{2}[\log \,(1+{{x}^{2}})]\,_{0}^{\infty }+\frac{1}{2}[{{\tan }^{-1}}x]\,_{0}^{\infty }\] \[=0+0+\frac{1}{2}\left[ \frac{\pi }{2}-0 \right]=\frac{\pi }{4}\].
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