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

1 Answer

Aug 2, 2016

#= -e^(-t)(t+1) + C#

Explanation:

For #u, v# functions of #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#

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