integration - Computing the integral $int_{-infty}^{infty}frac{z^4}{1+z^8}dz$

I need help to compute the following integral:

$$\int_{-\infty}^{\infty}\frac{z^4}{1+z^8}dz$$

I need to use Cauchy's residue theorem.

I can write that $z^8+1=z^8-i^2=(z^4-i)(z^4+i)$. How do I proceed?

Please give a methodological answer so that I can solve other questions too.

Thanks