What is the residue for the pole of $\frac{1}{az^4+bz^2+c}$

Question

I have to solve $\int_0^{\infty} \frac{1}{ax^4+bx^2+c} dx$ we can assume $(b^2 - 4ac \neq 0)$

I confirmed integral on arc with radius R converges to $0$ (when R -> $\infty$)

But I can't find the residue to use Residue theorem.

The calculation becomes too complex.

please tell me the residue and the way how to get them.

Thanks

edit:

$az^4+bz^2+c = 0$

$z = ±\sqrt{ \frac{-b ± \sqrt{b^2 - 4ac} }{2a}} $

$z_1 = +\sqrt{ \frac{-b + \sqrt{b^2 - 4ac} }{2a}}$

$z_2 = -\sqrt{ \frac{-b - \sqrt{b^2 - 4ac} }{2a}}$

$Res(z_1) = \lim_{z \to z_1}(z-z_1)\frac{1}{ax^4+bx^2+c}$

And I think this is too complex to solve. So I suspect that there is another way to get the residue.

Answer 1

Try to do a partial fraction decomposition with $\frac{1}{a(x^2-p)(x^2-q)}$.