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! 

These integrals are often wrapped up nicely by substitutions of the form:
$$u=x^a\pm\frac{1}{x^a}$$
where $a$ is chosen appropriately.
A little bit of playing around leads to the following:
$$\int\frac{x^{5}-x}{x^{8}+1}dx=\int\frac{x^{3}\left(x^{2}-\frac{1}{x^{2}}\right)dx}{x^{4}\left(x^{4}+\frac{1}{x^{4}}\right)}=\int\frac{\left(x^{2}-\frac{1}{x^{2}}\right)dx}{x\left[\left(x^{2}+\frac{1}{x^{2}}\right)^{2}-2\right]}$$
Now let
$$u=x^{2}+\frac{1}{x^{2}}$$
$$du=2\left(x-\frac{1}{x^{3}}\right)dx=2\frac{1}{x}\left(x^{2}-\frac{1}{x^{2}}\right)dx$$
Hence
$$2I=\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}}\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$
$$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x^{4}-\sqrt{2}x^{2}+1}{x^{4}+\sqrt{2}x^{2}+1}\right|$$
