Question Number 207857 by necx122 last updated on 28/May/24
$$\int{x}\mathrm{tan}^{−\mathrm{1}} {xdx} \\ $$
Answered by Ghisom last updated on 28/May/24
$$\mathrm{by}\:\mathrm{parts} \\ $$$$\int{x}\mathrm{arctan}\:{x}\:{dx}= \\ $$$$=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{arctan}\:{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\left({x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{arctan}\:{x}\:−{x}\right)+{C} \\ $$
Commented by necx122 last updated on 29/May/24
$${Thank}\:{you} \\ $$