A) ln 2
B) \[-\ln 2\]
C) \[\frac{\pi }{2}+\ln 2\]
D) \[\frac{\pi }{2}-\ln 2\]
Correct Answer: D
Solution :
\[\int_{0}^{1}{{{\tan }^{-1}}\left( \frac{1}{{{x}^{2}}-x+1} \right)\,dx}=\int_{0}^{1}{{{\tan }^{-1}}x\,dx-}\int_{0}^{1}{{{\tan }^{-1}}(x-1)}\,dx\] \[=2\int_{\,0}^{\,1}{{{\tan }^{-1}}x\,dx}=2\,[{{\tan }^{-1}}x-\frac{1}{2}\log (1+{{x}^{2}})]_{0}^{1}=\frac{\pi }{2}-\log 2.\]You need to login to perform this action.
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