A) \[\log 2\]
B) \[\frac{\pi }{2}+\log 2\]
C) \[\log 4\]
D) \[\frac{\pi }{2}-\log 4\]
Correct Answer: A
Solution :
\[2\int\limits_{0}^{1}{{{\tan }^{-1}}}xdx=\int\limits_{0}^{1}{{{\cot }^{-1}}}(1-x+{{x}^{2}})dx\] ?(1) \[\int\limits_{0}^{1}{{{\tan }^{-1}}}(1-x+{{x}^{2}})dx=\int\limits_{0}^{1}{\left\{ \frac{\pi }{2}-{{\cot }^{-1}}(1-x+{{x}^{2}})dx \right\}}\]\[=\frac{\pi x}{2}\left| _{0}^{1} \right.-2\int\limits_{{}}^{{}}{{{\tan }^{-1}}xdx=\frac{\pi }{2}-2\left( \frac{\pi }{4}-\frac{1}{2}\ell n2 \right)}=\ell n2\]
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