Question Number 155897 by cortano last updated on 05/Oct/21
$$\:\int\:\mathrm{x}\left(\mathrm{arctan}\:\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{dx}=? \\ $$
Answered by puissant last updated on 05/Oct/21
$${Q}=\int{x}\left({arctanx}\right)^{\mathrm{2}} {dx}.. \\ $$$${IBP}\:\:\Rightarrow\:.. \\ $$$$\begin{cases}{{u}={arctan}\:{x}}\\{{v}'={xarctan}\:{x}}\end{cases}\:\:\Rightarrow\:\:\begin{cases}{{u}'=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }}\\{{v}=−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctanx}}\end{cases} \\ $$$$\Rightarrow\:{Q}=−\frac{{x}}{\mathrm{2}}{arctanx}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctan}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\int\left\{{arctanx}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right\}{dx} \\ $$$$=−\frac{{x}}{\mathrm{2}}{arctanx}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctan}^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{2}}\left({xarctanx}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\because\:\:\:{Q}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){arctanx}−{x}\:{arctanx}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{C}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:………………………\mathscr{L}{e}\:{puissant}…………………… \\ $$
Commented by SANOGO last updated on 06/Oct/21
$${le}\:{dur}\:{gar} \\ $$