A) \[\pi /4\]
B) \[{{\pi }^{2}}/32\]
C) 1
D) none of these
Correct Answer: B
Solution :
\[\int_{0}^{\pi }{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}\]put \[{{\tan }^{-1}}x=t\] \[\frac{1}{1+{{x}^{2}}}dx=dt\] \[\int_{0}^{\pi /4}{t\,dt=\left[ \frac{{{t}^{2}}}{2} \right]}_{0}^{\pi /4}={{\pi }^{2}}/32\]You need to login to perform this action.
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