Answer
$\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$
Work Step by Step
The integration by parts formula suggests that $\int a'(x) b(x)=a(x) b(x)-\int a(x) b'(x)dx$
Now, we have
$I=\int x^2 \ln x dx$
$\implies I=(\dfrac{x^3}{3})(\ln x)-\int (\dfrac{x^3}{3}) (\dfrac{1}{x})dx$
This implies that
$I=(\dfrac{x^3}{3})(\ln x)-\int (\dfrac{x^2}{3})dx$
$\implies I=\dfrac{x^3}{3}\ln x-(\dfrac{x^3}{9})+c$
