Answer
$$\frac{1}{3}x^3{ {{{\tan }^{ - 1}}x}}{} - \frac{{{x^2}}}{6} + \frac{1}{6}\ln \left( {{x^2} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}{{\tan }^{ - 1}}x} dx \cr
& {\text{Integrating by parts}}{\text{, }} \cr
& {\text{Let }}u = {\tan ^{ - 1}}x,{\text{ }}du = \frac{1}{{1 + {x^2}}}dx \cr
& {\text{ }}dv = {x^2}dx,{\text{ }}v = \frac{1}{3}{x^3} \cr
& {\text{Using integration by parts formula}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {{x^2}{{\tan }^{ - 1}}x} dx = { {{{\tan }^{ - 1}}x} }\left( {\frac{1}{3}{x^3}} \right) - \int {\left( {\frac{1}{3}{x^3}} \right)\left( {\frac{1}{{1 + {x^2}}}} \right)} dx \cr
& {\text{Multiply}} \cr
& \int {{x^2}{{\tan }^{ - 1}}x} dx = \frac{1}{3}{ {{{x^3\tan }^{ - 1}}x} }{} - \frac{1}{3}\int {\frac{{{x^3}}}{{{x^2} + 1}}} dx \cr
& {\text{Using the long division}} \cr
& = \frac{1}{3}{x^3 {{{\tan }^{ - 1}}x} }{} - \frac{1}{3}\int {\left( {x - \frac{x}{{{x^2} + 1}}} \right)} dx \cr
& = \frac{1}{3}{x^3 {{{\tan }^{ - 1}}x} }{} - \frac{1}{3}\int x dx + \frac{1}{3}\int {\frac{x}{{{x^2} + 1}}} dx \cr
& {\text{Integrate}} \cr
& = \frac{1}{3}{x^3 {{{\tan }^{ - 1}}x} }{} - \frac{1}{3}\left( {\frac{{{x^2}}}{2}} \right) + \frac{1}{3}\int {\frac{x}{{{x^2} + 1}}} dx \cr
& = \frac{1}{3}{x^3 {{{\tan }^{ - 1}}x} }{} - \frac{{{x^2}}}{6} + \frac{1}{6}\ln \left( {{x^2} + 1} \right) + C \cr} $$
