Answer
$$
\int \frac{1}{x^{7}-x} d x=\frac{1}{6} \ln \left|\frac{x^{6}-1}{x^{6}}\right|+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\begin{aligned} \int \frac{1}{x^{7}-x} d x &=\int \frac{d x}{x\left(x^{6}-1\right)}\\
&=\int \frac{x^{5}}{x^{6}\left(x^{6}-1\right)} d x \\
& \quad \quad \quad\left[\begin{array}{c}{ \text{Let }u=x^{6}} \\ {\text{ so that } d u=6 x^{5} d x}\end{array}\right] , \text{Then } \\
&=\frac{1}{6} \int \frac{1}{u(u-1)} d u \quad\quad \quad\quad(1) \end{aligned}
$$
Now, decompose $\frac{1}{u(u-1)}$ into its partial fractions as follows,
$$
\frac{1}{u(u-1)} =\frac{A}{u} + \frac{B}{(u-1)}
$$
$\Rightarrow \quad 1=A(u-1)+Bu= (A+B)u -A $ by comparison of the coefficients we get $A=-1, B=1.$
so $$
\frac{1}{u(u-1)} =\frac{-1}{u} + \frac{1}{(u-1)} \quad\quad (2)
$$
substituting from Eq. (2) into Eq. (1) we get
$$
\begin{aligned} \int \frac{1}{x^{7}-x} d x &=\frac{1}{6} \int \frac{1}{u(u-1)} d u\\
\\ &=\frac{1}{6} \int\left(\frac{1}{u-1}-\frac{1}{u}\right) d u\\
&=\frac{1}{6}(\ln |u-1|-\ln |u|)+C \\ &=\frac{1}{6} \ln \left|\frac{u-1}{u}\right|+C\\
&=\frac{1}{6} \ln \left|\frac{x^{6}-1}{x^{6}}\right|+C \end{aligned}
$$
Thus the given integral is:
$$
\int \frac{1}{x^{7}-x} d x=\frac{1}{6} \ln \left|\frac{x^{6}-1}{x^{6}}\right|+C
$$
