Show that

$\frac{1}{\left(e^{x}-1\right)}$

Let $e^{x}=t \Rightarrow e^{x} d x=d t$

$\Rightarrow \int \frac{1}{e^{x}-1} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$

Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$

$1=A(t-1)+B t$    ...(1)

Substituting t = 1 and t = 0 in equation (1), we obtain

A = −1 and B = 1

$\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$

$\begin{aligned} \Rightarrow \int \frac{1}{t(t-1)} d t &=\log \left|\frac{t-1}{t}\right|+\mathrm{C} \\ &=\log \left|\frac{e^{x}-1}{e^{x}}\right|+\mathrm{C} \end{aligned}$