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