# How do you integrate int te^-t by integration by parts method?

Aug 2, 2016

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

#### Explanation:

For $u , v$ functions of $t$,

$\int u v ' \mathrm{dt} = u v - \int u ' v \mathrm{dt}$

$u \left(t\right) = t \implies u ' \left(t\right) = 1$

$v ' \left(t\right) = {e}^{- t} \implies v \left(t\right) = - {e}^{- t}$

$\int t {e}^{- t} \mathrm{dt} = - t {e}^{- t} + \int {e}^{- t} \mathrm{dt}$

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