A) \[\frac{{{\pi }^{2}}}{8}\]
B) \[\frac{{{\pi }^{2}}}{16}\]
C) \[\frac{{{\pi }^{2}}}{4}\]
D) \[\frac{{{\pi }^{2}}}{32}\]
Correct Answer: D
Solution :
Put \[t={{\tan }^{-1}}x\Rightarrow dt=\frac{1}{1+{{x}^{2}}}dx,\] then \[\int_{0}^{1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx=\int_{0}^{\pi /4}{t\,dt=\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{\pi /4}=\frac{{{\pi }^{2}}}{32}}}\].
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