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xtan-1-xdx-




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

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