How do you integrate $\int t e^{- t}$ $int te^-t$ by integration by parts method?

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Answer 1 · 2016-08-02T14:38:40

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

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

$int uv'dt = uv - int u'vdt$

$u(t) = t implies u'(t) = 1$

$v'(t) = e^(-t) implies v(t) = -e^(-t)$

$intte^(-t)dt = -te^(-t) + int e^(-t)dt$

$=-te^(-t) - e^(-t) + C = -e^(-t)(t+1) + C$

How do you integrate int te^-t∫te−tint te^-t by integration by parts method?