# How do I find the integral int(x*e^-x)dx ?

Aug 4, 2014

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

Process:

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

This integral will require integration by parts. Keep in mind the formula:

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

We will let $u = x$, and $\mathrm{dv} = {e}^{- x} \mathrm{dx}$.

Therefore, $\mathrm{du} = \mathrm{dx}$. Finding $v$ will require a $u$-substitution; I will use the letter $q$ instead of $u$ since we are already using $u$ in the integration by parts formula.

$v = \int {e}^{- x} \mathrm{dx}$
let $q = - x$.

thus, $\mathrm{dq} = - \mathrm{dx}$

We will rewrite the integral, adding two negatives to accommodate $\mathrm{dq}$:

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

Written in terms of $q$:

$v = - \int {e}^{q} \mathrm{dq}$

Therefore,

$v = - {e}^{q}$

Substituting back for $q$ gives us:

$v = - {e}^{- x}$

Now, looking back at the IBP's formula, we have everything we need to start substituting:

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

Simplify, canceling the two negatives:

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

That second integral should be easy to solve - it's equal to $v$, which we've already found. Simply substitute, but remember to add the constant of integration:

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