Answer
\[ = \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{x^2}\ln x\,dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = \ln x\,\,\,\,\,\, \to \,\,\,\,\,du = \frac{{dx}}{x} \hfill \\
dv = {x^2}dx\,\,\, \to \,\,\,v = \frac{{{x^3}}}{3} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= \frac{{{x^3}\ln 3}}{3} - \int_{}^{} {\,\left( {\frac{{{x^3}}}{3}} \right)\,\left( {\frac{{dx}}{x}} \right)} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {{x^2}dx} \hfill \\
\hfill \\
integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C \hfill \\
\end{gathered} \]