Calculus: Early Transcendentals (2nd Edition)

$= \frac{1}{3}\,\,\left[ {{x^3}\,\,\ln \,{x^3} - {x^3}} \right] + C$
$\begin{gathered} \int_{}^{} {{x^2}} \ln {x^3}dx \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ {x^3} = t\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,3{x^2}dx = dt \hfill \\ {x^2}dx = \frac{{dt}}{3} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{1}{3}\int_{}^{} {\ln t\,\,dt} \hfill \\ \hfill \\ Take\,\,u = \ln t\,\, \to du = \frac{1}{t}\,\,dt \hfill \\ and\,\,\,dv = dt\, \to v = t \hfill \\ \hfill \\ integrating\,\,by\,\,parts \hfill \\ \hfill \\ use\,\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{1}{3}\,\,\left[ {t\,\,\,\ln \,\,t - \int_{}^{} t \,\,\frac{1}{t}\,\,dt} \right] \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = \frac{1}{3}\,\,\left[ {t\,\,\,\ln \,\,t - \int_{}^{} {dt} \,} \right] \hfill \\ = \frac{1}{3}\,\,\left[ {t\,\,\ln \,t - t} \right] + C \hfill \\ \hfill \\ substituting\,back\,:\,\,\,\,t = {x^3} \hfill \\ \hfill \\ = \frac{1}{3}\,\,\left[ {{x^3}\,\,\ln \,{x^3} - {x^3}} \right] + C \hfill \\ \hfill \\ \end{gathered}$