Could $\int\frac{1}{x^{N+1}(x-1)}dx$ be solved analytically?

Question

I am trying to solve this integral: $$\int\frac{1}{x^{N+1}(x-1)}dx$$
I have tried integration by partial fraction, substitution and by parts. But, I can't solve it. So, I would like to ask could this be solved?
Also, May I know when partial fraction does not exist?

Thank you very much.

Update: N is any number that is greater than 0. Sorry for forgetting to include such information, and apology for any inconvenience caused.

Answer 1

Hint:

Assuming that $N$ is natural, you may do the following

$$\int\frac{1}{x^{N+1}(x-1)}dx = \int\frac{1-x^{N+1}+x^{N+1}}{x^{N+1}(x-1)}dx$$ $$= -\int \frac 1{x^{N+1}}\sum_{n=0}^Nx^n \; dx + \int \frac{dx}{x-1}$$

The first sum comes from $$\frac 1{x^{N+1}}\frac{1-x^{N+1}}{x-1}=-\frac 1{x^{N+1}}\frac{x^{N+1}-1}{x-1}=-\frac 1{x^{N+1}}(1+x+\cdots + x^N)$$

Basically, the above is the partial fraction decomposition of the integrand.

Answer 2

If you try a few integer values of $N$, you'll notice that partial fraction decomposition gives you a pattern that suggests

$$\frac{1}{x^{N+1}(x-1)} = \frac{1}{x-1} - \sum_{k=1}^{N+1} \frac{1}{x^k}$$

From there, simply integrate both sides. They're fairly trivial integrals when rewritten as so.