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