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