Let $I=\int\limits \frac{1}{e^x-1}dx$ Put $e^x = t$ $\Rightarrow e^x dx = dt$ $\Rightarrow dx=\frac{dt}{t}$ $\therefore I=\int\limits \frac{dt}{t(t-1)}$ Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ $1 = A(t - 1) + Bt$ ...(i) Putting $t = 1$ in (i), we get $B = 1$ Putting $t = 0$ in (i), we get $1 = A(0 - 1) + B(0)$ $\Rightarrow A = -1$$\therefore \frac{1}{t(t-1)} = -\frac{1}{t} + \frac{1}{t-1}$$\therefore I = \int\limits \left( -\frac{1}{t} + \frac{1}{t-1} \right) dt$$= -\log|t| + \log|t-1| + C$$= \log\left(\frac{e^x-1}{e^x}\right) + C$