$x{\tan ^{ - 1}}x$

Let $I = \int {x{{\tan }^{ - 1}}x} dx$

$= {\tan ^{ - 1}}x \cdot \int x dx - \int {\left[ {\left( {\cfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right)} \right)\int {\left( {x\,dx} \right)} } \right]} dx$

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

$= \cfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \cfrac{1}{2}\int {\cfrac{{{x^2}}}{{{x^2} + 1}}dx} = \cfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \cfrac{1}{2}\int {\cfrac{{{x^2} + 1 - 1}}{{1 + {x^2}}}dx}$

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

$= \cfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \cfrac{1}{2}x + \cfrac{1}{2}{\tan ^{ - 1}}x + C$