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Question Number 207857 by necx122 last updated on 28/May/24

$$\int x{\tan }^{-1}xdx$$

Answered by Ghisom last updated on 28/May/24

$$\mathrm{by}\mathrm{parts}$$$$\int x\arctan xdx=$$$$=\frac{x^2}{2}\arctan x-\frac{1}{2}\int \frac{x^2}{x^2+1}dx=$$$$=\frac{1}{2}((x^2+1)\arctan x-x)+C$$

Commented by necx122 last updated on 29/May/24

$$Thankyou$$

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