# Integrate : x tan^-1 x / square root (1+x^2)

Solution:   We have

$I=\int \frac{x\tan^{-1}x}{\sqrt{1+x^{2}}}dx$

Integrating by parts  , we get

$u=\tan^{-1}x \Rightarrow du=\frac{dx}{1+x^{2}};v=\sqrt{1+x^{2}}\Rightarrow dv=\frac{xdx}{\sqrt{1+x^{2}}}\\ \\ \Rightarrow \hspace{1cm}I=\sqrt{1+x^{2}}\tan^{-1}x-\int \sqrt{1+x^{2}} \cdot \frac{dx}{1+x^{2}}\\ \\ \Rightarrow \hspace{1cm}I=\sqrt{1+x^{2}}\tan^{-1}x-\ln (x+\sqrt{x^{2}+1})+C$

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