How to integrate $\int 1/(x^7 -x) dx$? How should I proceed about this integral? 
$$\int {1/(x^7 -x)} dx$$
I've tried integration by parts or substitution but I can't seem to solve it. Can I have some hints on how should I get started?
These are some of the things I've tried:
IBP: $u = \frac {1}{x^6-1}$, $du = \frac {-5x^6}{x^6-1}$, $dv = \frac 1x dx$, $v = \ln|x|$
Tried substitution method, but not successful.
 A: 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.
A: As in this example it is useful to extract something of the form
$$u=x^a\pm\frac{1}{x^a}$$
$$\int\frac{dx}{x^{7}-x}=\int\frac{1}{x^{4}}\frac{dx}{x^{3}-\frac{1}{x^{3}}}=\int\frac{\left(x^{2}+\frac{1}{x^{4}}\right)dx}{x^{3}-\frac{1}{x^{3}}}-\int\frac{x^{2}dx}{x^{3}-\frac{1}{x^{3}}}=I_1-I_2$$
Now in $I_1$ we may let
$$u=x^{3}-\frac{1}{x^{3}}$$
$$du=3\left(x^{2}+\frac{1}{x^{4}}\right)dx$$
So that
$$I_1=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}\ln|u|=\frac{1}{6}\ln u^2=\frac{1}{6}\ln\frac{(x^6-1)^2}{x^6}$$
In $I_2$ apparently $v=x^3$
$$I_2=\frac{1}{3}\int\frac{vdv}{v^{2}-1}=\frac{1}{6}\ln\left(v^{2}-1\right)=\frac{1}{6}\ln|x^6-1|$$
Finally,
$$I=I_1-I_2=\frac{1}{6}\ln\frac{x^6-1}{x^6}=\frac{1}{6}\ln\left(1-\frac{1}{x^6}\right)$$
A: Alternatively the fastest method is to simply observe the following 
$$\int \frac{1}{x^7-x}dx  = \int \frac{1}{x^7(1-x^{-6})}dx = \int \frac{x^{-7}}{1-x^{-6}}dx = \frac{1}{6}ln(1-x^{-6})$$
A: 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.
A: The cheater's method is to "observe" that
$$
\frac{d}{dx} \ln\left(1-\frac{1}{x^6}\right) = \frac{6}{x^7-x}
$$
