How to integrate $\int 1/(x^7 -x) dx$?

One way to look at the problem is to say that it would be easy if the integrand were $$\frac{7x^6-1}{x^7-x},$$and also easy if it were $$\frac{x^6-1}{x^7-x}.$$Now take a linear combination of these to knock out the $x^6$ term in the numerator.

The cheater's method is to "observe" that $$ \frac{d}{dx} \ln\left(1-\frac{1}{x^6}\right) = \frac{6}{x^7-x} $$

Partial fractions: $x^7-x=x(x^6-1)=x(x^3-1)(x^3+1)$, and then use the standard factorizations of $a^3-b^3$ and $a^3+b^3$ to split each of the cubic factors into a linear factor and a quadratic factor.