# How do you integrate (lnx)^2/x?

Notice that $\left(\ln x\right) ' = \frac{1}{x}$ hence
$\int {\left(\ln x\right)}^{2} / x \mathrm{dx} = \int {\left(\ln x\right)}^{2} \cdot \left(\ln x\right) ' \mathrm{dx} = \frac{1}{3} \cdot {\left(\ln x\right)}^{3} + c$