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

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

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

$$=\int \frac{tg\left(s\right)s^2}{{(1+{tg}^2\left(s\right))}^{\frac{1}{2}}}dss={tan}^{-}\left(x\right)$$$$=\int s^{2}sin\left(s\right)dseasytoofindnow$$$$$$

Commented by jagoll last updated on 14/Feb/20

$$whatistgsir?\tan ?$$

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

$$yestangent$$

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

$$changementx=tan\theta give$$$$I=\int \frac{tan\theta \times {\theta }^2}{{(1+{tan}^2\theta )}^{\frac{3}{2}}}(1+{tan}^2\theta )d\theta =\int \frac{{\theta }^{2}tan\theta }{\sqrt{1+{tan}^2\theta }}$$$$I=\int {\theta }^{2}tan\theta cos\theta d\theta =\int {\theta }^{2}sin\theta d\theta {=}_{byparts}$$$$-{\theta }^{2}cos\theta +\int 2\theta cos\theta d\theta =-{\theta }^{2}cos\theta +2\{\theta sin\theta -\int sin\theta \}$$$$=-{\theta }^{2}cos\theta +2\theta sin\theta +2cos\theta =(2-{\theta }^2)cos\theta +2\theta sin\theta +C$$$$=(2-arctanx)cos\left(artanx\right)+2arctan\left(x\right)sin\left(arctanx\right)+C$$$$=\frac{2-arctanx}{\sqrt{1+x^2}}+\frac{2xarctanx}{\sqrt{1+x^2}}+C$$

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