Answer
\[ = \frac{{{x^2}{{\tan }^{ - 1}}{x^2}}}{2} - \frac{1}{4}\ln \,\left( {1 + {x^4}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {x{{\tan }^{ - 1}}\,\left( {{x^2}} \right)dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = {\tan ^{ - 1}}{x^2}\,\,\,\,\,\,\,\, \to \,\,\,du = \frac{{2x}}{{1 + {x^4}}}dx \hfill \\
du = xdx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,u = \frac{{{x^2}}}{2} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{Therefore}}{\text{,}} \hfill \\
\hfill \\
\frac{{{x^2}{{\tan }^{ - 1}}{x^2}}}{2} - \int_{}^{} {\,\left( {\frac{{{x^2}}}{2}} \right)\,\left( {\frac{{2x}}{{1 + {x^4}}}} \right)} dx \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{{x^2}{{\tan }^{ - 1}}{x^2}}}{2} - \int_{}^{} {\frac{{{x^3}}}{{1 + {x^4}}}} \,\,dx \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \frac{{{x^2}{{\tan }^{ - 1}}{x^2}}}{2} - \frac{1}{4}\int_{}^{} {\frac{{4{x^3}}}{{1 + {x^4}}}} \,\,dx \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{{{x^2}{{\tan }^{ - 1}}{x^2}}}{2} - \frac{1}{4}\ln \,\left( {1 + {x^4}} \right) + C \hfill \\
\end{gathered} \]
