Calculating $\sum_{n=1}^\infty {\frac{nx^n}{4n^2-1}}$

Question

I would appreciate any help calculating the series. And determine where does the series converge uniformly.

$$\sum_{n=1}^\infty {\frac{nx^n}{4n^2-1}} $$

Answer 1

Hint. You may write $$ \frac{nx^n}{4n^2-1}=\frac{x^n}{4 (2 n-1)}+\frac{x^n}{4 (2 n+1)} $$ and one may recall

$$ \sum_{n=0}^\infty\frac{u^{2n+1}}{2 n+1}=\frac12 \log\left(\frac{1+u}{1-u}\right), \qquad |u|<1. $$