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$\int_{0}^{1} x \tan^{- 1} x d x$

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Correct option is A

∫01 xtan−1⁡xdx=x22⋅tan−1⁡x01−∫01 11+x2⋅x22⋅dx=π8−12∫01 1−11+x2dx=π8−12x−tan−1⁡x01=π8−121−π4−0=π4−12

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∫01xtan−1x dx

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$\int_{0}^{1} x \tan^{- 1} x d x$