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Question
Solution
\[Let\ I = \int_0^1 x \tan^{- 1} x\ d\ x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \frac{x^2}{1 + x^2} dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \int_0^1 \left( \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \right) dx\]
\[ \Rightarrow I = \left[ \frac{x^2 \tan^{- 1} x}{2} \right]_0^1 - \frac{1}{2} \left[ x - \tan^{- 1} x \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{8} - 0 - \frac{1}{2}\left( 1 - \frac{\pi}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \frac{1}{2}\]
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