Solution
= Solution
Use <integration by parts> with $u=\ln x$ and $dv=x^2\,dx$. This gives the <antiderivative>
$$
\int x^2\ln x\,dx=\frac{x^3}{3}\ln x-\frac13\int x^2\,dx
=\frac{x^3}{3}\ln x-\frac{x^3}{9}+C.
$$
Evaluating at the two bounds, with $\ln1=0$, yields
$$
\boxed{\int_1^y x^2\ln x\,dx=\frac{y^3}{3}\ln y-\frac{y^3-1}{9}.}
$$