Integrate $e^{f(x)}$ Just wondering how I can integrate $\displaystyle xe^{ \large {-x^2/(2\sigma^2)}}$
Tried using substitution where $U(x) = x^2$ but I kept getting a $x^2$ at the denominator which is incorrect.
I understand that $\displaystyle \int e^{f(x)} = e^{\large \frac{f(x)}{f'(x)}}$ if $f(x)$ is linear, however, how do we handle this situation when $f(x)$ is not linear?
Step by step answer will be awesome!!!! thanks :D
M

A close relative of your substitution, namely $u=-\dfrac{x^2}{2\sigma^2}$, works. 
In "differential" notation, we get $du=-\frac{1}{\sigma^2} x\,dx$, so $x\,dx=-\sigma^2\,du$.  
Remark: Or more informally, let's guess that the answer is $e^{-x^2/(2\sigma^2)}$. Differentiate, using the Chain Rule. We get $-\frac{x}{\sigma^2}e^{-x^2/(2\sigma^2)}$, so wrong guess. Too bad.
But we got close, and there is an easy fix by multiplying by a suitable constant to get rid of the $-\frac{1}{\sigma^2}$ in front of the derivative of our wrong guess. The indefinite integral is $-\sigma^2e^{-x^2/(2\sigma^2)}+C$.  
