Question
Integrate the rational function $\frac{1}{\left(e^{x}-1\right)}$ [Hint: ex = t]

Answer

Given function is, $\frac{1}{\left(e^{x}-1\right)}$ 
Let ex = t
ex dx = dt
$\int \frac{1}{\left(e^{x}-1\right)} d x=\int \frac{1}{t-1} \times \frac{d t}{t}=\int \frac{1}{t(t-1)} d t$ 
Let $\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}$ 
1 = A(t - 1) + Bt … (i)
Substituting t =1 and t = 0 in equation (i), we get,
A = -1 and B = 1
Therefore, $\frac{1}{t(t-1)}=\frac{-1}{t}+\frac{1}{t-1}$ 
$\int \frac{1}{t(t-1)} d t=\log \left|\frac{t-1}{t}\right|+C$ 
= $\log \left|\frac{e^{x}-1}{e^{x}}\right|+C$

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