# Integrate xe^x dx

• Sep 3rd 2009, 02:16 PM
jebckr
Integrate xe^x dx
it is actually integrate from 0 to -infin, but I understand that part. When I see xe^x I integrate it and get (1/2x^2)(e^x). My solutions manuel says it is (x-1)(e^x). I don't get the same answer when using my work, what am I doing wrong?
• Sep 3rd 2009, 02:22 PM
Matt Westwood
Integrate by parts: $\int u dv = uv - \int v du$.

Here you set $u = x, dv = e^x$ and so $uv = x e^x$ and $\int v du = \int e^x = e^x$.

Hence, as they say, the result.
• Sep 3rd 2009, 02:29 PM
jebckr
So how does (xe^x)-(e^x) become (x-1)e^x? I'm sure I'm missing some simple rule here...
• Sep 3rd 2009, 02:35 PM
cgiulz
Yes:

(xe^x)-(e^x)

= xe^x - e^x

= e^x(x - 1)

= (x - 1)e^x