How do you integrate int xe^-x by integration by parts method?

Nov 27, 2016

$- {e}^{-} x \left(x + 1\right) + C$

Explanation:

We have the integral $\int x {e}^{-} x \mathrm{dx}$. We want to apply integration by parts, which fits the form $\int u \mathrm{dv} = u v - \int \mathrm{dv} u$. So for the given integral, let:

$\left\{\begin{matrix}u = x & \implies & \mathrm{du} = \mathrm{dx} \\ \mathrm{dv} = {e}^{-} x \mathrm{dx} & \implies & v = - {e}^{-} x\end{matrix}\right.$

To go from $\mathrm{dv}$ to $v$, the integration will require a substitution. Think for $\int {e}^{-} x \mathrm{dx}$, let $t = - x$.

So:

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

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