# How do you integrate int xe^xdx using integration by parts?

Remember the formula for IBP: $\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$
Let $u = x \implies \frac{\mathrm{du}}{\mathrm{dx}} = 1$
Let $\frac{\mathrm{dv}}{\mathrm{dx}} = {e}^{x} \implies v = {e}^{x}$
$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \left(1\right) \mathrm{dx}$
$\therefore \int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \mathrm{dx}$
$\therefore \int x {e}^{x} \mathrm{dx} = x {e}^{x} - {e}^{x} + c$