How do you use integration by parts to find intxe^x dx?

1 Answer
Oct 7, 2014

To integrate by parts, we have to pick a $u$ and $\mathrm{dv}$ such that

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

We can pick what $u$ and $\mathrm{dv}$ are, though we should typically try to pick $u$ such that $\mathrm{du}$ is "simpler". (By the way, $\mathrm{du}$ just means the derivative of $u$, and $\mathrm{dv}$ just means derivative of $v$)

So, from $\int u \mathrm{dv}$, let's pick $u = x$ and $\mathrm{dv} = {e}^{x}$. It is helpful to fill in all of the parts you will use throughout the problem before you start integrating:

$u = x$
$\frac{\mathrm{du}}{\mathrm{dx}} = 1$ so $\mathrm{du} = 1 \mathrm{dx} = \mathrm{dx}$

$\mathrm{dv} = {e}^{x}$
$v = \int \mathrm{dv} = \int {e}^{x} \mathrm{dx} = {e}^{x}$

Now let's plug everything back into the original formula.

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$
$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \mathrm{dx}$
You can integrate the last part quite simply now, to get:

$\int x {e}^{x} \mathrm{dx} = x {e}^{x} - \int {e}^{x} \mathrm{dx} = x {e}^{x} - {e}^{x} + c$