A) \[-\frac{\pi }{2}-1\]
B) \[\frac{\pi }{2}+1\]
C) \[\frac{\pi }{2}\]
D) None of these
Correct Answer: A
Solution :
\[\int_{-1}^{1}{x{{\tan }^{-1}}x\,dx}=\left[ \left\{ {{\tan }^{-1}}x.\frac{{{x}^{2}}}{2} \right\} \right.\] \[\left. -\int_{-1}^{1}{\frac{1}{1+{{x}^{2}}}.\frac{{{x}^{2}}}{2}dx} \right]\] \[=\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int{\frac{{{x}^{2}}+1-1}{1+{{x}^{2}}}dx} \right]_{-1}^{1}\] \[=\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}\int{\left( 1-\frac{1}{1+{{x}^{2}}} \right)dx} \right]_{-1}^{1}\] \[=\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}(x-{{\tan }^{-1}}x) \right]_{-1}^{1}\] \[=\left[ \frac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\frac{1}{2}x+\frac{{{\tan }^{-1}}x}{2} \right]_{-1}^{1}\] \[=\frac{1+1}{2}{{\tan }^{-1}}(1)-\frac{1}{2}-\frac{1+1}{2}{{\tan }^{-1}}(-1)-\frac{1}{2}\] \[={{\tan }^{-1}}\left( \tan \frac{\pi }{4} \right)-{{\tan }^{-1}}\left( \tan \frac{3\pi }{4} \right)-1\] \[=\frac{\pi }{4}-\frac{3\pi }{4}-1\] \[=\frac{-2\pi }{4}-1\] \[=-\frac{\pi }{2}-1\]
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