Help for evaluating complicated integral $\int \frac 1 {x^n-x} dx$

Hint :

Let I=$\large\int \dfrac 1 {x^n-x} dx$

I=$\large\int \dfrac 1{x} \cdot \dfrac 1 {x^{n-1}-1} dx$

I=$\large\int \dfrac {x^{n-2}}{x^{n-1}} \cdot \dfrac 1 {x^{n-1}-1} dx$

let $\large x^{n-1}=t$ $\implies dt=\big(n-1)x^{n-2}\ dx$

so, I=$\large \dfrac{1}{n-1}\cdot \int \dfrac {dt}{t\cdot (t-1)} $

use partial fractions now..

You're done!!

$$\int\dfrac{1}{x^n-x}\mathrm{d}x=\int\dfrac{1}{x^n\left(1-\dfrac{1}{x^{n-1}}\right)}\mathrm{d}x$$

Now subs. $\dfrac{1}{x^{n-1}}=t\implies \mathrm{d}t=-\dfrac{(n-1)}{x^n}\mathrm{d}x$

Hence your integral becomes $$\dfrac{1}{(n-1)}\int\dfrac{\mathrm{d}t}{t-1}$$