How do you integrate

(x-r)^2 (1/r) (e^(-x/r)) from the interval negative infinity to positive infinity?

let

u = (x-r)^2
du = 2(x-r) dx

dv = (1/r) e^(-x/r) dx
v = e^(-x/r)

∫ u dv = uv - ∫ v du
= (x-r)^2 e^(-x/r) - 2∫(x-r) e^(-x/r) dx

Now do it all over again. Notice that the power of (x-r) has been reduced by one.