Answer
$\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int x^2 \ln x dx=\dfrac{x^3}{3}\ln x-\int (\dfrac{x^3}{3}) (\dfrac{1}{x})dx$
Thus, $\dfrac{x^3}{3}\ln x-\int (\dfrac{x^2}{3})dx=\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$