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! 
 A: You may try using factorization $x^8+1=(x^4-\sqrt{2}{\ }x^2+1)(x^4+\sqrt{2}{\ }x^2+1)$
A: 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
$$2I=\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}}\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$
$$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x^{4}-\sqrt{2}x^{2}+1}{x^{4}+\sqrt{2}x^{2}+1}\right|$$
A: Decomposition in partial fractions. Easy calculation.
http://en.wikipedia.org/wiki/Partial_fraction
A: In real quadratics,
$ x^8 + 1$ is  $$ (x^2 - \; x \sqrt{2 + \sqrt 2} \; + 1) (x^2 +  x \sqrt{2 + \sqrt 2}  + 1) (x^2 -  x \sqrt{2 - \sqrt 2}  + 1) (x^2 +  x \sqrt{2 - \sqrt 2}  + 1)  $$
I got this by finding $\cos \frac{\pi}{8}$ and $\sin \frac{\pi}{8}.$ On the other hand, you can easily see the relationship with the answer of M. Strochyk.
What this means is that, at the cost of square roots all over creation, partial fractions can be carried out completely, most likely including $\arctan$ and $\log$ terms, what ever else usually comes up.
A: Letting $y=x^2$ reduces the powers of $x$ and then dividing both the denominator and numerator of the integrand by $y^2$ yields
$$
\begin{aligned}
\int \frac{x^5-x}{x^8+1} d x&=\frac{1}{2} \int \frac{y^2-1}{y^4+1} d y \\
=& \frac{1}{2} \int \frac{1-\frac{1}{y^2}}{y^2+\frac{1}{y^2}} d y \\
=& \frac{1}{2} \int \frac{d\left(y+\frac{1}{y}\right)}{\left(y+\frac{1}{y}\right)^2-2} \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{y+\frac{1}{y}-\sqrt{2}}{y+\frac{1}{y}+\sqrt{2}}\right|+C \\
=& \frac{1}{4 \sqrt{2}} \ln \left|\frac{x^4-\sqrt{2} x^2+1}{x^4+\sqrt{2} x^2+1}\right|+C
\end{aligned}
$$
