Evaluate $\int_C\frac{dw}{e^w-1}$ (counterclockwise) over some loop C contained in the annulus $0<|z|<2\pi$.
Considering the coefficient of $1/z$ in the Laurent series for $\frac{1}{e^z-1}$ by long-dividing $\frac{1}{z}(\frac{z}{z+z^2/2!+z^3/3!+...})$, I get 1, which means the contour integral is $2\pi i$. Is there a direct way to evaluate the integral? The antiderivative of $\frac{1}{e^z-1}$ is $\log(e^z-1)-z+c$, but $\log$ isn't analytic throughout the disk, so the integral isn't necessarily zero.
Let's start computing the indefinite integral as you proposed : $$\int_C \frac {dw}{e^w-1}=\int_C \frac {-e^{-w}}{e^{-w}-1}dw=\log(e^{-w}-1)|_C$$ At this point you have to be careful and use appropriate starting and ending points for $C$ (so that no discontinuity appears during evaluation of $\ f(w):=\log(e^{-w}-1)\ $ on the path).
You may use the principal branch of the complex logarithm that excludes the branch cut $(-\infty,0]$. Here since $e^{-w}-1$ reverts the sign of real values we will have to exclude the interval $[0,\infty)$ from the values allowed for $w$.
Your path should start in your annulus over $\ w=r\ $ (with $0<r<2\pi$), turn counterclockwise around the origin from angle $0$ (excluded) up to angle $2\pi$.
The value of the integral will thus be :
(the definition of the principal value makes in fact unnecessary the limit for the left term i.e. at $2\pi$) \begin{align} L&=\lim_{\epsilon\to0^+}\left[\operatorname{Log}(e^{-(r-i\epsilon)}-1)-\operatorname{Log}(e^{-(r+i\epsilon)}-1)\right]\\ &=\operatorname{Log}(1-e^{-r})+\pi i-\left(\operatorname{Log}(1-e^{-r})-\pi i\right)\\ &=2\pi i\\ \end{align}