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

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 $$2I=\int\frac{du}{u^{2}-2}=\frac{1}{2\sqrt{2}}\int\frac{du}{u-\sqrt{2}}-\int\frac{du}{u+\sqrt{2}}=\frac{1}{2\sqrt{2}}\ln\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|$$ $$I=\frac{1}{4\sqrt{2}}\ln\left|\frac{x^{4}-\sqrt{2}x^{2}+1}{x^{4}+\sqrt{2}x^{2}+1}\right|$$

You may try using factorization $x^8+1=(x^4-\sqrt{2}{\ }x^2+1)(x^4+\sqrt{2}{\ }x^2+1)$