Answer
$$\frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}x + \frac{1}{2}{\tan ^{ - 1}}x + C$$
Work Step by Step
$$\eqalign{
& \int {x{{\tan }^{ - 1}}x} dx \cr
& {\text{substitute }}u = {\tan ^{ - 1}}x,{\text{ }}du = \frac{1}{{1 + {x^2}}}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& {\text{applying integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& = \left( {{{\tan }^{ - 1}}x} \right)\left( {\frac{{{x^2}}}{2}} \right) - \int {\left( {\frac{{{x^2}}}{2}} \right)\left( {\frac{1}{{1 + {x^2}}}} \right)dx} \cr
& {\text{multiply}} \cr
& = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} \cr
& {\text{long division}} \cr
& = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {1 - \frac{1}{{1 + {x^2}}}} \right)} dx \cr
& = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx \cr
& {\text{find the antiderivative}} \cr
& = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}x + \frac{1}{2}{\tan ^{ - 1}}x + C \cr} $$