Answer
$$\frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{{{x^2}}}{3} + \frac{4}{3}\ln \left( {{x^2} + 4} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx \cr
& {\text{Using the integration by parts method }} \cr
& \,\,\,\,\,{\text{Let }}\,\,\,\,\,u = {\tan ^{ - 1}}\frac{x}{2},\,\,\,\,du = \frac{{1/2}}{{1 + {{\left( {x/2} \right)}^2}}}\,\,\,dx,\,\,\,\,du = \frac{2}{{4 + {x^2}}}dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {x^2}dx,\,\,\,\,v = \frac{1}{3}{x^3} \cr
& \cr
& {\text{Integration by parts formula: }}\int {udv} = uv - \int {vdu.{\text{ T}}} {\text{hen}}{\text{,}} \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \int {\frac{1}{3}{x^3}} \left( {\frac{2}{{4 + {x^2}}}} \right)\,\,\,dx \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{2}{3}\int {\frac{{{x^3}}}{{4 + {x^2}}}} \,dx \cr
& \cr
& {\text{Use long division for }}\frac{{{x^3}}}{{4 + {x^2}}} \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{2}{3}\int {\left( {x - \frac{{4x}}{{{x^2} + 4}}} \right)} \,dx \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{2}{3}\int x \,dx + \frac{4}{3}\int {\frac{{2x}}{{{x^2} + 4}}} \,dx \cr
& \cr
& {\text{integrating}} \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{2}{3}\left( {\frac{{{x^2}}}{2}} \right) + \frac{8}{3}\ln \left( {{x^2} + 4} \right) + C \cr
& \cr
& {\text{factoring}} \cr
& \int {{x^2}} {\tan ^{ - 1}}\frac{x}{2}dx = \frac{1}{3}{x^3}{\tan ^{ - 1}}\frac{x}{2} - \frac{{{x^2}}}{3} + \frac{8}{3}\ln \left( {{x^2} + 4} \right) + C \cr} $$
