<s>[INST] Evaluate: $\int \frac{1}{x^7-x}\ \mathrm{d}x$ Evaluate:
$$\int \frac{1}{x^7-x}\ \mathrm{d}x$$
My approach to this question:
$$\int \frac{1}{x^7-x}\ \mathrm{d}x = \int \frac{1}{x(x^6-1)}\ \mathrm{d}x$$
$$\int \frac{1}{x(x^6-1)}\ \mathrm{d}x = \int \frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)}\ \mathrm{d}x$$
$$\frac{1}{x(x-1)(x+1)(x^2-x+1)(x^2+x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} + \frac{Dx+E}{x^2-x+1} + \frac{Fx+G}{x^2+x+1}$$
At this point I realized how brutal this question if going to be. Is there an easier way to solve the integral?
 [/INST] Yes there is. Note that:
$$\frac{1}{x^7-x} =\frac{x^5}{x^6-1}-\frac{1}{x}$$ 
Now use $t = x^6-1$ for the first integral, and...
Note how this trick works for any integral of the form $(x^n-x)^{-1}$.
</s>