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Question Number 81636 by jagoll last updated on 14/Feb/20

∫ ((x(tan^(−1) (x))^2 )/((1+x^2 )^(3/2) )) dx =

$$\int\:\frac{{x}\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dx}\:=\: \\ $$

Commented by mind is power last updated on 14/Feb/20

=∫((tg(s)s^2 )/((1+tg^2 (s))^(1/2) ))ds   s=tan^− (x)  =∫s^2 sin(s)ds easy too find now

$$=\int\frac{{tg}\left({s}\right){s}^{\mathrm{2}} }{\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({s}\right)\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{ds}\:\:\:{s}={tan}^{−} \left({x}\right) \\ $$$$=\int{s}^{\mathrm{2}} {sin}\left({s}\right){ds}\:{easy}\:{too}\:{find}\:{now} \\ $$$$ \\ $$

Commented by jagoll last updated on 14/Feb/20

what is tg sir? tan ?

$${what}\:{is}\:{tg}\:{sir}?\:\mathrm{tan}\:? \\ $$

Commented by peter frank last updated on 14/Feb/20

yes tangent

$${yes}\:{tangent} \\ $$

Commented by mathmax by abdo last updated on 14/Feb/20

changement x=tanθ give   I =∫((tanθ ×θ^2 )/((1+tan^2 θ)^(3/2) ))(1+tan^2 θ)dθ =∫ ((θ^2  tanθ)/(√(1+tan^2 θ)))  I=∫ θ^2 tanθ cosθ dθ =∫ θ^2 sinθ dθ  =_(byparts)   −θ^2 cosθ +∫  2θ cosθ dθ =−θ^2  cosθ +2 {  θ sinθ −∫ sinθ}  =−θ^2  cosθ +2θ sinθ +2cosθ =(2−θ^2 )cosθ +2θ sinθ +C  =(2−arctanx)cos(artanx)+2 arctan(x)sin(arctanx) +C  =((2−arctanx)/(√(1+x^2 ))) +((2x arctanx)/(√(1+x^2 ))) +C

$${changement}\:{x}={tan}\theta\:{give}\: \\ $$$${I}\:=\int\frac{{tan}\theta\:×\theta^{\mathrm{2}} }{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\int\:\frac{\theta^{\mathrm{2}} \:{tan}\theta}{\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}} \\ $$$${I}=\int\:\theta^{\mathrm{2}} {tan}\theta\:{cos}\theta\:{d}\theta\:=\int\:\theta^{\mathrm{2}} {sin}\theta\:{d}\theta\:\:=_{{byparts}} \\ $$$$−\theta^{\mathrm{2}} {cos}\theta\:+\int\:\:\mathrm{2}\theta\:{cos}\theta\:{d}\theta\:=−\theta^{\mathrm{2}} \:{cos}\theta\:+\mathrm{2}\:\left\{\:\:\theta\:{sin}\theta\:−\int\:{sin}\theta\right\} \\ $$$$=−\theta^{\mathrm{2}} \:{cos}\theta\:+\mathrm{2}\theta\:{sin}\theta\:+\mathrm{2}{cos}\theta\:=\left(\mathrm{2}−\theta^{\mathrm{2}} \right){cos}\theta\:+\mathrm{2}\theta\:{sin}\theta\:+{C} \\ $$$$=\left(\mathrm{2}−{arctanx}\right){cos}\left({artanx}\right)+\mathrm{2}\:{arctan}\left({x}\right){sin}\left({arctanx}\right)\:+{C} \\ $$$$=\frac{\mathrm{2}−{arctanx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+\frac{\mathrm{2}{x}\:{arctanx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:+{C} \\ $$

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