Evaluate integral - Wolfram Alpha works with a summation in terms of $\omega$ I found this "monster"
$$\int_0^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx$$
and I would like to know how to evaluate it. If not a complete answer, at least some hints.
Wolfram Alpha (here) works with a summation in terms of $\omega$. Can somebody explain me the meaning of this?

With @Grant B.’s guide and Wolfram Alpha , I obtained the roots.
I will try to make the partial fractions during the week-end.
 A: $\lim_{\omega \to \infty}$, use partial fraction and instead of inserting infinity,insert $\omega$ , you will get same result.
A: This is too long for a comment.
As you wrote, this is a real monster and I wish you very pleasant days with partial fraction decomposition. In any manner, since the roots of $x^7+6x^4+9=0$ cannot be expressed but just numerically evaluated, you could not get any "exact" result.
What I would suggest is just numerical integration $$\int_0^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx=\int_0^{1}\frac{x^5−6x^2}{x^7+6x^4+9}dx+\int_1^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx$$ For the second integral, use $x=\frac 1y$ to get $$\int_1^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx=-\int_0^{1}\frac{6 y^3-1}{9 y^7+6 y^3+1}dy$$
Edit
By the end, we need to compute $$I=\int_0^{1}\left(\frac{x^5−6x^2}{x^7+6x^4+9}- \frac{6 x^3-1}{9 x^7+6 x^3+1}\right)\,dx$$ Using the trapezoidal method with $101$ points, we then end with $$I=0.0837499$$ while the "exact" answer would be $0.0837447$.
Since all of that can easily be done using Excel, repeat the calculations using $1001$ points and get $0.0837447$ which is the exact solution for six significant figures.
