Taylor series of fraction addition of common function What is the typical trick for finding the taylor series of a common function that is in the denominator when adding a constant. 
eg:
$$f(x)=\frac{1}{e^x-c}$$
I know you can write $f(x)=\frac{e^x}{e^{2x}-ce^x}$ and then invoke 
$$Taylor(\frac{f(x)}{g(x)})=\frac{Taylor{f(x)}}{Taylor(g(x))}$$
but I feel that there might be an easier way to evaluate $f(x)$
Any hint would be appreciated
Edit:
the origin of this question is to do integrals like
$$\int_a^b (\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kt}}-1})\,d\lambda$$  
and limits for the same function
$$\lim_{\lambda \to +\lambda_0}{\frac{8\pi hc}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kt}}-1}}$$

If the problem for you is that the function is in the denominator, why not just take it as the numerator, i.e. re-write
$f(x) = \frac{1}{e^{x} - 2}$
as
$f(x) = (e^{x} - 2)^{-1}$
and proceed as
$f'(x) = -(e^{x} -2)^{-2}\cdot e^{x} \\
f''(x) = 2(e^{x} -2)^{-3}\cdot e^{2x} - (e^{x} -2)^{-2}\cdot e^{x} \\
f'''(x) = -6(e^{x}-2)^{-4}\cdot e^{3x} + 4(e^{x} -2)^{-3}\cdot e^{2x} + 2(e^{x} -2)^{-3}\cdot e^{2x} - (e^{x} -2)^{-2}\cdot e^{x} \\
\ldots$
where you should be able to write the derivatives more compactly
