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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

∫x^2 ((tan^(−1) x)/(x^2 +1))dx  =∫tan^(−1) (x)−∫((tan^(−1) (x))/(x^2 +1))dx  =xtan^(−1) (x)−∫(x/(1+x^2 ))dx−(tan^(−1) (x))^2   =xtan^(−1) (x)−(1/2)log(1+x^2 )−(tan^(−1) (x))^2 +C

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

Commented by aliibrahim1 last updated on 27/Apr/21

thx man

$${thx}\:{man} \\ $$

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