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Question Number 197744 by mathlove last updated on 27/Sep/23

$$2{\int }_0^1{tan}^{-1}xdx=?$$

Answered by witcher3 last updated on 27/Sep/23

$$\int {2\tan }^{-1}\left(\mathrm{x}\right)\mathrm{dx}={2\mathrm{xtan}}^{-1}\left(\mathrm{x}\right)-\int \frac{2\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\mathrm{xtan}}^{-1}\left(\mathrm{x}\right)-\ln (1+{\mathrm{x}}^2)+\mathrm{c},\mathrm{c}\in \mathbb{R}$$

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