Find the $E[X^3]$ of the normal distribution Find the $E[X^3]$ of the normal distribution with mean μ and variance $σ^2$ (in terms of $μ$ and $σ$).

So far, I have that it is the integral of $x^3$ multiplied with the pdf of the normal distribution, but when I try to integrate it by parts, it becomes super convulated especially with the e term. I know there's tricks with odd and even functions that may apply and I have a sinking suspicion it might just end up being μ, but I've kind of hit a wall. Any help would be appreciated :) 

EDIT:

So I think the answer is either $0$ (because of how the graph looks when I plotted it) or, by simple integration it is $3μσ^2 + μ^3$.
You can observe that it is a integral of a impar function over the all domain $\mathbb R$. Therefore it is equal to zero
Try replacing $x^2$ by $t$ and do the integration by parts.