Answer
$$\int x^{2}lnx\,dx=\frac{x^{3}lnx}{3}-\frac{x^{3}}{9}+C$$
Work Step by Step
$${(\frac{x^{3}lnx}{3})}'=x^{2}lnx+\frac{x^{2}}{3}$$
$$\int x^{2}lnx\,dx=\frac{x^{3}lnx}{3}-\frac{1}{3}\int x^{2}dx$$
$$=\frac{x^{3}lnx}{3}-\frac{x^{3}}{9}+C$$