# How do you integrate int x^2lnx by integration by parts method?

Use integration by parts. Let $u = \ln x$ and $\mathrm{dv} = {x}^{2} \mathrm{dx}$ Then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = \frac{1}{3} {x}^{3}$.
$\int {x}^{2} \ln x \mathrm{dx} = \frac{1}{3} {x}^{3} \ln x - \int \left(\frac{1}{3} {x}^{3} \cdot \frac{1}{x}\right) \mathrm{dx}$
$\int {x}^{2} \ln x \mathrm{dx} = \frac{1}{3} {x}^{3} \ln x - \int \frac{1}{3} {x}^{2} \mathrm{dx}$
$\int {x}^{2} \ln x \mathrm{dx} = \frac{1}{3} {x}^{3} \ln x - \frac{1}{9} {x}^{3} + C$