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Question Number 43624 by peter frank last updated on 12/Sep/18

$$\underset{0}{\stackrel{1}{\int }}\frac{{\tan }^{-1}x}{1+x^2}dx=$$

Commented by math khazana by abdo last updated on 13/Sep/18

$$letI={\int }_0^1\frac{arctanx}{1+x^2}dxletintegratebyparts$$$$u^{{}^{\prime }}=\frac{1}{1+x^2}and{v}^{}=arctanx\Rightarrow$$$$I={\left[{arctan}^{2}x\right]}_0^1-{\int }_0^1\frac{arctanx}{1+x^2}dx=\frac{{\pi }^2}{16}-I\Rightarrow$$$$2I=\frac{{\pi }^2}{16}\Rightarrow I=\frac{{\pi }^2}{32}.$$

Answered by username last updated on 13/Sep/18

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