What's the sum function of $\sum_{n=1}^{\infty} \frac{x^n(x-1)}{n(x^{2n+1}-1)}$? Notice that 
\begin{align*}
\frac{x^n(x-1)}{n(x^{2n+1}-1)}&=\frac{1}{n}\cdot\frac{x^n}{\sum_{k=0}^{2n}x^k}=\frac{1}{n\sum_{k=0}^{2n}x^{k-n}}=\frac{1}{n\sum_{k=-n}^{n}x^{k}}
\end{align*}
This will help?

What about
\begin{align*}
(1-x)\sum_{n=1}^\infty \frac{1}{n} \frac{x^n}{1-x^{2n+1}} = (1-x) \sum_{m=0}^\infty \sum_{n=1}^\infty \frac{1}{n} x^{n + (2n +1)m} = (x-1) \sum_{m=0}^\infty x^m \ln(1-x^{2m+1}),
\end{align*}
where we used the series representation of the logarithm.
But I am not sure, if it is possible to give a closed formula for the last series.
