A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{8}\]
C) \[\frac{{{\pi }^{2}}}{16}\]
D) \[\frac{{{\pi }^{2}}}{32}\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}}dx\] Let \[{{\tan }^{-1}}x=t\] \[\Rightarrow \] \[\frac{1}{1+{{x}^{2}}}dx=dt\] when \[x=0,\text{ }t=0\] and \[x=1,t=\frac{\pi }{4}\] \[\therefore \] \[I=\int_{0}^{\pi /4}{t\,dt}\] \[=\left( \frac{{{t}^{2}}}{2} \right)_{0}^{\frac{\pi }{4}}=\frac{{{\pi }^{2}}}{16\times 2}\] \[=\frac{{{\pi }^{2}}}{32}\]
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