Help integrate $\frac{1}{x^3+x^8}$ Could you help me to integrate 
$$
\int{\frac{dx}{x^3+x^8}}
$$
I've tried partial fraction decomposition but got $(x^4-x^3+x^2-x+1)$ as the last term when factored the denominator. 
Thank you.

The best way to get the answer given by Wolfram Alpha is to write:
$$\frac{1}{x^3+x^8} = \frac{1}{x^3} -\frac{x^2}{x^5+1}$$
Then solve:
$$\frac{x^2}{1+x^5}=\frac{a}{x+1} + \frac{bx+c}{x^2-2\cos(\pi/5)x+1}+\frac{dx+e}{x^2-2\cos(3\pi/5)x+1}$$
Since $\sin\pi/5$ and $\cos\pi/5$ are in terms of $\sqrt{5}$, you can rewrite the answer in terms of $\sin(\pi/5),\cos(\pi/5),\sin(3\pi/5),\cos(3\pi/5)$. It still won't be prtty.
