$$\int \frac {e^x}{(e^x - 1)} dx = \ln|e^x - 1| + C .....A $$ $$\int-\frac {e^x}{(1- e^x)}dx = \ln|1- e^x| + C ......B$$
$$\text{for A: } e^x > 1 \implies e^x > e^0 \implies x>0 $$
$$\text{for B: } e^x < 1 \implies e^x < e^0 \implies x<0$$
These integration occur inside larger questions, since solution of x varies drastically, which one to follow?
Note the absolute values inside the $\ln$ expressions: this means that you don't need to restrict to $x > 0$ or $x < 0$, only $x \neq 0$.
Keep in mind that the constants may be different on either side of $x=0$, however, so to be careful, we should have:
$$\int \frac {e^x}{(e^x - 1)} dx = \begin{cases} \ln(e^x - 1) + C_1, & \text{if $x > 0$}\\ \ln(1 - e^x) + C_2, & \text{if $x < 0$}\\ \end{cases} $$
$$\frac{-e^x}{1-e^x}=\frac{-1}{-1}\cdot\frac{e^x}{-1+e^x}=\frac{e^x}{e^x-1}$$
$$\ln\left\lvert 1-e^x\right\rvert = \ln\left\lvert (-1)\left(-1+e^x\right)\right\rvert = \ln\left( \lvert-1\rvert\cdot\left\lvert-1+e^x\right\rvert\right)=\ln\left( 1\cdot\left\lvert -1+e^x\right\rvert\right)=\ln\left\lvert e^x-1\right\rvert$$
Does that help? Here is a graph.
