How do I find the maclaurin representation for $f(x)=\frac{\ln(2x+1)}{e^x-1}$? $f(x)=\frac{\ln(2x+1)}{e^x-1}$
I'm supposed to find the Maclaurin representation for $f(x)$ and then write down the first 4 terms. The question says I should use known the Maclaurin representation of elemental functions. Thus, I've tried using $e^x=\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\Rightarrow e^x-1=\displaystyle\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\right)-1$ and using $\ln(2x+1)=\displaystyle\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^n}{n}$.
I've also tried inserting my function in a Maclaurin Series Representation calculator and I got:
$\frac{\ln(2x+1)}{e^x−1}≈\frac{3103x^4}{360}−\frac{11x^3}{2}+\frac{23x^2}{6}−3x+2$
I've also tried finding the first 4 derivatives by hand and replacing $x=0$. When I do that I find that they're all equal to zero.
Thank you to anyone willing to give me a hand.

This is  a problem of composition of series
$$\log(2x+1)=2 x-2 x^2+\frac{8 x^3}{3}-4 x^4+\frac{32 x^5}{5}-\frac{32 x^6}{3}+\frac{128
   x^7}{7}-32 x^8+O\left(x^9\right)$$
$$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{
   x^7}{5040}+\frac{x^8}{40320}+O\left(x^9\right)$$ Now long division
$$f(x)=\frac{\ln(2x+1)}{e^x-1}=\frac{2 x-2 x^2+\frac{8 x^3}{3}-4 x^4+\frac{32 x^5}{5}-\frac{32 x^6}{3}+\frac{128
   x^7}{7}-32 x^8+O\left(x^9\right) } {x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{
   x^7}{5040}+\frac{x^8}{40320}+O\left(x^9\right)}$$
$$f(x)=2-3 x+\frac{23 x^2}{6}-\frac{11 x^3}{2}+\frac{3103 x^4}{360}-\frac{5111
   x^5}{360}+\frac{365129 x^6}{15120}-\frac{635437 x^7}{15120}+O\left(x^8\right)$$ that you can truncate whereever you wish.
