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

Question

I have this complicated integral to evaluate :

$$\int \dfrac 1 {x^n-x} dx$$

I'm struggling to evaluate this.

My attempt :

$$\int \dfrac1x \cdot \dfrac 1 {x^{n-1}-1} dx$$ Now, I try to apply integration by parts. For that, I use : $V=\large\dfrac1x$ and $U=\large\dfrac 1 {x^{n-1}-1}$

But, that doesnt take me anywhere. It just gives me an even more complicated expression to evaluate..

Help would be appreciated.. Thank you..

Answer 1

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!!

Answer 2

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

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

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

Answer 3

$$ \begin{aligned} \int \frac{1}{x^n-x} & d x=\int \frac{1}{x\left(x^{n-1}-1\right)} d x \\ & =\int\left(\frac{x^{n-2}}{x^{n-1}-1}-\frac{1}{x}\right) d x \\ & =\frac{1}{n-1} \int \frac{d\left(x^{n-1}\right)}{x^{n-1}-1}-\int \frac{1}{x} d x \\ & =\frac{1}{n-1} \ln \left|x^{n-1}-1\right|-\ln |x|+C \end{aligned} $$