# How do you integrate int x^2 ln 5x dx  using integration by parts?

Start by letting $u = \ln \left(5 x\right)$ and $\mathrm{dv} = {x}^{2} \setminus \mathrm{dx}$ to ultimately get $\int \setminus {x}^{2} \ln \left(5 x\right) \setminus \mathrm{dx} = \frac{1}{3} {x}^{3} \ln \left(5 x\right) - {x}^{3} / 9 + C$.
If you let $u = \ln \left(5 x\right)$ and $\mathrm{dv} = {x}^{2} \setminus \mathrm{dx}$, then $\mathrm{du} = \frac{1}{5 x} \cdot 5 \setminus \mathrm{dx} = \frac{1}{x} \setminus \mathrm{dx}$ and $v = {x}^{3} / 3$. Therefore
$\setminus \int \setminus {x}^{2} \ln \left(5 x\right) \setminus \mathrm{dx} = u v - \setminus \int \setminus v \setminus \mathrm{du}$
$= \frac{1}{3} {x}^{3} \ln \left(5 x\right) - \setminus \int \setminus {x}^{2} / 3 \setminus \mathrm{dx} = \frac{1}{3} {x}^{3} \ln \left(5 x\right) - {x}^{3} / 9 + C$.