Question Number 35681 by prof Abdo imad last updated on 22/May/18
$${find}\:\:\int\:{arctan}\left({x}\right){dx} \\ $$
Commented by prof Abdo imad last updated on 23/May/18
$${let}\:{integrate}\:{by}\:{parts}\:{u}^{'} \:=\mathrm{1}\:{and}\:{v}={arctan}\left({x}\right) \\ $$$${I}\:=\:{x}\:{arctanx}\:\:−\int\:\:\:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}\:=\:{xarctsnx}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+\lambda\:\:\:\:\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18
![∫tan^(−1) xdx =tan^(−1) x∫dx−∫[((dtan^(−1) x)/dx)∫dx]dx =xtan^(−1) x−∫(x/(1+x^2 ))dx =xtan^(−1) x−(1/2)ln(1+x^2 )](https://www.tinkutara.com/question/Q35711.png)
$$\int{tan}^{−\mathrm{1}} {xdx} \\ $$$$={tan}^{−\mathrm{1}} {x}\int{dx}−\int\left[\frac{{dtan}^{−\mathrm{1}} {x}}{{dx}}\int{dx}\right]{dx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xtan}^{−\mathrm{1}} {x}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right) \\ $$