# Whats the integral of xe^(-x)

Whats the integral of $x{e}^{-x}$
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losnonamern
In order to use substitution, one part of the integrand must be the derivative of the other part...that is not the case here. to integrate the above equation requires integration by parts.
$\int x{e}^{-x}dx$
$\int udv=uv-\int vdu$
let:
u=x and $dv={e}^{-x}dx$
then
du=dx and $v=-{e}^{-x}$
plugging in.
$\int x{e}^{-x}dx=-x{e}^{-x}dx=-x{e}^{-x}-{e}^{-x}+C=-{e}^{-x}\left(x+1\right)+C$
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Makena Preston
integration by parts:
let u= x $dv={e}^{-x}dx$
du= dx $v=-{e}^{-x}$
so $uv-\int vdu=x\ast \left(-{e}^{-x}\right)-\int \left(-{e}^{-x}\right)dx$
$=-x{e}^{-x}+\int {e}^{-x}dx$ [use u-sub to solve integral: let $u=-x⇒du=-dx$]
$=-x{e}^{-x}-{e}^{-x}+C$