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

`int xe^(-x) dx = -xe^(-x) - e^(-x) + C`

Process:

`int x e^(-x) dx =` ?

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

`int u dv = uv - int v du`

We will let `u = x`, and `dv = e^(-x) dx`.

Therefore, `du = 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) dx`
let `q = -x`.

thus, `dq = -dx`

We will rewrite the integral, adding two negatives to accommodate `dq`:

`v = -int -e^(-x) dx`

Written in terms of `q`:

`v = -int e^(q) 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 xe^(-x) dx = x*(-e^(-x)) - int -e^(-x) dx`

Simplify, canceling the two negatives:

`int xe^(-x) dx = -xe^(-x) + int e^(-x) 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 xe^(-x) dx = -xe^(-x) - e^(-x) + C`