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Question Number 89918 by jagoll last updated on 20/Apr/20

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

Commented by john santu last updated on 20/Apr/20

$$[u={\tan }^{-1}\left(x\right)\Rightarrow du=\frac{dx}{1+x^2}]$$$$v=\int \frac{xdx}{{(1+x^2)}^{3/2}}=\frac{1}{2}\int \frac{d(1+x^2)}{{(1+x^2)}^{3/2}}$$$$v=-\frac{1}{\sqrt{1+x^2}}]$$$$\Rightarrow I=-\frac{{\tan }^{-1}\left(x\right)}{\sqrt{1+x^2}}+\int \frac{dx}{{(1+x^2)}^{3/2}}$$$$letJ=\int \frac{dx}{{(1+x^2)}^{3/2}}$$$$x=\tan p\Rightarrow dx=\sec {}^{2}pdp$$$$J=\int \frac{\sec {}^{2}pdp}{\sec {}^{3}p}=\int \cos pdp$$$$J=\sin p=\frac{x}{\sqrt{1+x^2}}$$$$thereforeweget$$$$\int \frac{x{\tan }^{-1}\left(x\right)dx}{{(1+x^2)}^{3/2}}=-\frac{{\tan }^{-1}\left(x\right)}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}+c$$$$=\frac{x-{\tan }^{-1}\left(x\right)}{\sqrt{1+x^2}}+c$$

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