Another interesting integral $\int{\frac{x^m}{x^{2m}-1}dx}$ After An interesting integral $\int{\dfrac{x^m}{x^{2m}+1}dx}$, I want to generalize a similar integral below
$$\int{\frac{x^m}{x^{2m}-1}dx}$$
for all values of $m\in\mathbb{N}$. Below are my steps:
$$\int{\frac{x^m}{x^{2m}-1}dx}=\int{\frac{x^m}{(x^m-1)(x^m+1)}dx}=\frac{1}{2}\left(\int{\frac{1}{x^m-1}dx}+\int{\frac{1}{x^m+1}dx}\right)$$
I don't know what my next step should be.
How should I solve this integral?

Fractionalize  partially the integrand
$$\frac{x^n}{x^{2n}-1}=\sum_{k=1}^{2n}\frac{c_k}{x-x_k}$$
with $x_k= e^{i a_k},\>a_k=\frac{k\pi}{n}$ and apply the L’Hoptital’s rule to obtain the coefficients
$$c_k = \lim_{x\to x_k}\frac{x^n(x-x_k)}{x^{2n}-1}=\frac{(-1)^kx_k}{2n}$$
Then
\begin{align}
&\int \frac{x^n}{x^{2n}-1}dx =\frac1{2n}\sum_{k=1}^{2n}\int \frac{(-1)^kx_k}{x-x_k} dx 
= \frac1{2n}\sum_{k=1}^{2n} (-1)^kx_k\ln(x-x_k)\\
= &\frac1{2n} \sum_{k=1}^{2n}(-1)^k (\cos a_k +i \sin a_k)
\ln(x-\cos a_k -i\sin a_k)\\
= &\frac1{2n} \sum_{k=1}^{2n} (-1)^k\left(\cos a_k \ln\sqrt{x^2-2x\cos a_k+1}+\sin a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k} \right)
\end{align}
