A) \[\left( \frac{\pi }{2}-1 \right)\]
B) \[\left( \frac{\pi }{2}+1 \right)\]
C) \[(\pi -1)\]
D) 0
Correct Answer: A
Solution :
\[I=\int_{-1}^{1}{x{{\tan }^{-1}}x\,dx=2}\int_{0}^{1}{x{{\tan }^{-1}}x\,dx}\] \[\because \,\,\,x{{\tan }^{-1}}x\]is an even function \[I=[2\frac{{{x}^{2}}}{2}{{\tan }^{-1}}x]_{0}^{1}-2\int_{0}^{1}{\frac{1}{2}\frac{{{x}^{2}}}{1+{{x}^{2}}}dx}\] \[I=[{{x}^{2}}{{\tan }^{-1}}x]_{0}^{1}-\int_{0}^{1}{\frac{{{x}^{2}}+1-1}{1+{{x}^{2}}}dx}\] I =\[[{{x}^{2}}{{\tan }^{-1}}x]_{0}^{1}-[x]_{0}^{1}+[{{\tan }^{-1}}x]_{0}^{1}\] Þ \[I=\frac{\pi }{4}-1+\frac{\pi }{4}=\frac{\pi }{2}-1\].
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