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Question Number 139454 by aliibrahim1 last updated on 27/Apr/21

Answered by Dwaipayan Shikari last updated on 27/Apr/21

$$\int x^2\frac{{tan}^{-1}x}{x^2+1}dx$$$$=\int {tan}^{-1}\left(x\right)-\int \frac{{tan}^{-1}\left(x\right)}{x^2+1}dx$$$$={xtan}^{-1}\left(x\right)-\int \frac{x}{1+x^2}dx-{\left({tan}^{-1}\left(x\right)\right)}^2$$$$={xtan}^{-1}\left(x\right)-\frac{1}{2}log(1+x^2)-{\left({tan}^{-1}\left(x\right)\right)}^2+C$$

Commented by aliibrahim1 last updated on 27/Apr/21

$$thxman$$

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