How find the poles/residues of $\int_{-\infty}^\infty \frac{x^2 \, dx}{1 + x^4}$ I'm trying to find the poles/residues of this integral:
$$\int_{-\infty}^\infty \frac{x^2 \, dx}{1 + x^4}$$
I've been given this attempt for a solution, but I don't really understand the procedure for how the poles are found. 
EDIT:
From the solution on the bottom of this post:
Specifically I don't see how: 
$ \frac{p(z)}{q'(z)} = \frac{z^2}{4z^3} = \frac{1}{4z} .$
becomes this:
$ \frac{p(z)}{q'(z)} = - \frac{z^3}{4} $
What exactly is happening here?
END EDIT
As I understand it there should be $2$ poles, each of order $1$.




Hint:


*

*Fist find the roots of $z^4 = -1$. 

*Choose a path containg $2$ of those roots. Say a semi-circunference of radius $R > 1$. 

*Use the following Theorem: 



If $p$ and $q$ are functions, analytic at $z_0$, with $p(z_0) \neq 0$ and $q(z_0) = 0$ and $q'(z_0) \neq 0$ then $z_0$ is a simple pole of $p(z)/q(z)$ and $$\mathrm{Res}_{z=z_0} \frac{p(z)}{q(z)}= \frac{p(z_0)}{q'(z_0)}$$

what are $p$ and $q$?
