Calculus
posted by Christina .
I tried to integrate this:
intergral sign with an interval from inifinity to + infinity
function: (xr)^2 (1/r) (e^(x/r))
with respect to dx.....
I know that Steve set up the usubstitution but I seem to never arrive at an answer... It seems like I keep integrating and integrating and integrating and never arrive at an answer!!!!!!!!

Steve gave you:
let
u = (xr)^2
du = 2(xr) dx
dv = (1/r) e^(x/r) dx
v = e^(x/r)
CHECK SIGN
∫ u dv = uv  ∫ v du
= (xr)^2 e^(x/r)  2∫(xr) e^(x/r) dx
now the second part of it. He said to do it again.
u = x  r
du = dx
dv = e^x/r
v = r e^x/r
(xr)e^x/r  ∫re^x/r dx
now you can do ∫re^x/r dx = (1/r)e^x/r
CHECK THE SIGNS !
then put it back together and do the limits 
I got r^2  2r.. and it is wrong.... please help....

See whether you can get wolframalpha's result:
http://www.wolframalpha.com/input/?i=%E2%88%AB+%28%28xr%29^2+%281%2Fr%29+%28e^%28x%2Fr%29%29%29+dx
The limit would be undefined for x > ∞
I suspect a typo somewhere. Seems like there should be e^(x^2/r) or something.
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