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

Answer


$\begin{array}{l} I =\int \frac{d x}{\left(e^x-1\right)} \\ =\int \frac{e^x}{e^x\left(e^x-1\right)} d x \\ \rightarrow \text { Take, } e^x=t \\ \therefore e^x \cdot d x=d t \\ \text { I }=\int \frac{d t}{t(t-1)} \\ \text { I }=\int \frac{t-(t-1)}{t(t-1)} d t \\ =\int \frac{d t}{t-1}-\int \frac{d t}{t} \\ =\log |t-1|-\log |t|+c\end{array}$
$\begin{array}{l} I =\log \left|e^x-1\right|-\log \left|e^x\right|+c \\ I =\log \left|\frac{e^x-1}{e^x}\right|+c\end{array}$

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