Answer
$$\frac{1}{2}\left( {{x^2} + 1} \right){\tan ^{ - 1}}x - \frac{1}{2}x + C$$
Work Step by Step
$$\eqalign{
& \int x {\tan ^{ - 1}}xdx \cr
& {\text{Using integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\tan ^{ - 1}}x,\,\,\,\,du = \frac{1}{{1 + {x^2}}}\,\,\,dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = xdx,\,\,\,\,v = \frac{1}{2}{x^2} \cr
& \cr
& {\text{Integration by parts formula }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \int {\frac{1}{2}{x^2}} \left( {\frac{1}{{1 + {x^2}}}} \right)\,\,\,dx \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}} dx \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {\frac{{1 + {x^2}}}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}}} \right)} dx \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\left( {1 - \frac{1}{{1 + {x^2}}}} \right)} dx \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}\int {dx} + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx \cr
& {\text{integrating}} \cr
& \int x {\tan ^{ - 1}}xdx = \frac{1}{2}{x^2}{\tan ^{ - 1}}x - \frac{1}{2}x + \frac{1}{2}{\tan ^{ - 1}}x + C \cr
& {\text{factoring}} \cr
& = \left( {\frac{1}{2}{x^2}{{\tan }^{ - 1}}x + \frac{1}{2}{{\tan }^{ - 1}}x} \right) - \frac{1}{2}x + C \cr
& = \frac{1}{2}\left( {{x^2} + 1} \right){\tan ^{ - 1}}x - \frac{1}{2}x + C \cr} $$
