Question Number 47852 by maxmathsup by imad last updated on 15/Nov/18
$$\left.\mathrm{1}\right)\:{find}\:\int\:{x}\:{arctan}\left({x}\right){dx} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{arctan}\left({x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 16/Nov/18
![1) let A =∫ x arctan(x)dx by parts A =(x^2 /2)arctan(x)−∫ (x^2 /2) (dx/(1+x^2 )) =(x^2 /2)arctan(x)−(1/2) ∫ ((x^2 +1−1)/(x^2 +1))dx =(x^2 /2)arctan(x)−(x/2) +(1/2) arctan(x)+c . 2) ∫_0 ^1 x arctan(x)dx =[(x^2 /2)arctan(x)−(x/2) +((arctan(x))/2)]_0 ^1 =(π/8) −(1/2) +(π/8) =(π/4) −(1/2) .](https://www.tinkutara.com/question/Q47923.png)
$$\left.\mathrm{1}\right)\:{let}\:{A}\:=\int\:\:{x}\:{arctan}\left({x}\right){dx}\:{by}\:{parts}\: \\ $$$${A}\:=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}\right)−\int\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}\right)−\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:{arctan}\left({x}\right)+{c}\:. \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{arctan}\left({x}\right){dx}\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{arctan}\left({x}\right)−\frac{{x}}{\mathrm{2}}\:+\frac{{arctan}\left({x}\right)}{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\pi}{\mathrm{8}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{8}}\:=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$