## Calculus: Early Transcendentals (2nd Edition)

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