Evaluating $\int \frac{1}{x^{7} - x} ~ d{x} $. How do I evaluate the following indefinite integral?
$$
\int \frac{1}{x^{7} - x} ~ d{x}.
$$
Could someone give me some advice as to what method I should use or the steps that I should take?

Note: The OP originally requested for help in evaluating $ \displaystyle \int \left( \frac{1}{x^{7}} - x \right) ~ d{x} $, which may not have been his/her actual intention.
 A: There is a trick.  We have
$$\frac{1}{x^7-x}=\frac{7x^6}{x^7-x} -\frac{7x^6-1}{x^7-x}.$$ 
The first function is $\dfrac{7x^5}{x^6-1}$. For integrating, there is an obvious substitution. 
For the second function, there already is an obvious substitution. 
One can invent many examples that yield to the same sort of trick. 
A: $\displaystyle \int\frac{1}{x^7-x}dx = \int\frac{1}{x^7.\left(1-\frac{1}{x^6}\right)}dx$
Put $\displaystyle \left(1-\frac{1}{x^6}\right) = t$ and $\displaystyle \frac{6}{x^7}dx = dt$
$\displaystyle = \frac{1}{6}\int\frac{1}{t}dt = \frac{1}{6}\ln \mid t \mid+C$
$\displaystyle = \frac{1}{6}\ln \left|\frac{x^6-1}{x^6}\right|+C$
A: $$
\begin{aligned}
\int \frac{1}{x^{7}-x} d x &=\int \frac{1}{x\left(x^{6}-1\right)} d x \\
&=\int\left(\frac{x^{5}}{x^{6}-1}-\frac{1}{x}\right) d x \\
&=\int \frac{x^{5} d x}{x^{6}-1}-\ln |x| \\
&=\frac{1}{6} \ln \left|x^{6}-1\right|-\ln |x|+C \\
&=\frac{1}{6} \ln \left|1-\frac{1}{x^{6}}\right|+C .
\end{aligned}
$$
Replacing 6 by $n$ and changing the sign yields the general integral
$$\int \frac{1}{x\left(x^{n}\pm1\right)} d x=\mp \frac{1}{n} \ln \left|1\pm\frac{1}{x^{n}}\right|+C\tag*{}$$
A: Factor the denom into $x{(x^3-1)}(x^{3}+1)$ and do a u-sub $x^{3}+1=u$ and it will go easy.
