Odd, I don't even know whether it's neither or not

Given the function:

$\begin{aligned} f(x) & = \frac x{e^x-1} + \frac x2 + 1 & \small \color{#3D99F6} \text{Replacing }x \text{ with }-x \\ f(-x) & = \frac {-x}{e^{-x}-1} - \frac x2 + 1 \\ & = \frac {-xe^x}{1-e^x} - \frac x2 + 1 \\ & = \frac {xe^x}{e^x-1} - \frac x2 + 1 \\ & = \frac {xe^x-x+x}{e^x-1} - \frac x2 + 1 \\ & = x + \frac x{e^x-1} - \frac x2 + 1 \\ & = \frac x{e^x-1} + \frac x2 + 1 \end{aligned}$

This implies that $f(-x) = f(x)$ , there $f(x)$ is even ,