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

Jan 9, 2016

$I = {e}^{x} \left(x - 1\right) + c$

#### Explanation:

$I = \int x {e}^{x} \mathrm{dx}$

Say $u = x$ so $\mathrm{du} = 1$, $\mathrm{dv} = v = {e}^{x}$

$I = x {e}^{x} - \int {e}^{x} \mathrm{dx}$
$I = x {e}^{x} - {e}^{x}$
$I = {e}^{x} \left(x - 1\right) + c$