Question

Integrate the rational function $\frac{1}{(e^x-1)}$

Answer 1

$\frac{1}{(e^x-1)}$
Let $e^x=t\Rightarrow e^{x}dx=dt$
$\Rightarrow \int \frac{1}{e^x-1}dx=\int \frac{1}{t-1}\times \frac{dt}{t}=\int \frac{1}{t(t-1)}dt$
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$
$1=A(t-1)+Bt$ ............ (1)
Substituting $t=1$ and $t=0$ in equation (1), we obtain
$A=-1$ and $B=1$
$\therefore \frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$
$\Rightarrow \int \frac{1}{t(t-1)}dt=\log \left|\frac{t-1}{t}\right|+C$
$=\log \left|\frac{e^x-1}{e^x}\right|+C$