Indefinite integral of $\frac1{1-x^a}$

Question

Is there any known result for the following indefinite integral for any $a>0$ ? $$ \int \frac{1}{1-x^a}\,\mathrm{d}x $$

For $a=1 ~\& ~2$, I could get expressions via different methods(change of variables and fraction separation respectively). But can we tell anything for any general $a$?

Thanks in advance for any help in this regard!

Answer 1

Since the $n$-th derivative of $z^\alpha$ is given by $$D^nz^\alpha=p(\alpha,n)z^{\alpha-n}$$ Where $p(\alpha,n)=\prod_{k=1}^{n}(\alpha-k+1)$, we have $$D^nz^{\alpha}|_{z=1}=p(\alpha,n)$$ Because of this, we get the Taylor series for $z^\alpha$: $$z^\alpha=\sum_{n\geq 0}p(\alpha,n)\frac{(z-1)^n}{n!}$$ Plugging in $z=1-x^a$, and $\alpha=-1$, we get $$\frac{1}{1-x^a}=\sum_{n\geq0}\frac{p(-1,n)}{n!}(1-x^a-1)^n$$ $$\frac{1}{1-x^a}=\sum_{n\geq0}(-1)^n p(-1,n)\frac{x^{an}}{n!}$$ Which can be integrated term-by-term.

Answer 2

When $a$ is a positive integer, $1-x^a$ always factors into real linear and quadratic factors. Then you can use fraction separation, and it ends up a sum of logarithms and arc-tangents. Ask Wolfram Alpha for the first few.