Evaluate: $\int{\frac{x^{5}-x}{x^{8}+1}}\:\mathrm dx.$ 
Evaluate: $$\int{\frac{x^{5}-x}{x^{8}+1}\:\mathrm dx}.$$

I am unable to see a decent starting point for this integral, there are no radicals so trigonometric substitution isn't helpful; there is no nice partial fraction decomposition to simplify the integrand, integration by parts doesn't help to simplify it much, and I cannot see any factorization or useful substitution to use.
Can anyone help shed some light on this integral?
Thanks in advance! 

Letting $y=x^2$ reduces the powers of $x$ and then dividing both the denominator and numerator of the integrand by $y^2$ yields
$$
\begin{aligned}
\int \frac{x^5-x}{x^8+1} d x&=\frac{1}{2} \int \frac{y^2-1}{y^4+1} d y \\
=& \frac{1}{2} \int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}} d y \\
=& \frac{1}{2} \int \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^2-2} \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{y+\frac{1}{y}-\sqrt{2}}{y+\frac{1}{y}+\sqrt{2}}\right|+C \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{x^4-\sqrt{2} x^2+1}{x^4+\sqrt{2} x^2+1}\right|+C
\end{aligned}
$$
