$\cfrac{1}{{\left( {{e^x} - 1} \right)}}$ [Hint : Put ${e^x} = t$ ]

.: Let $I = \int {\cfrac{1}{{{e^x} - 1}}} dx$

Put ${e^x} = t$ $\Rightarrow$ ${e^x}dx = dt$ $\Rightarrow$ $dx = \cfrac{{dt}}{t}$

$\therefore$ $I = \int {\cfrac{{dt}}{{t\left( {t - 1} \right)}}}$

We write, $\cfrac{1}{{t\left( {t - 1} \right)}} = \cfrac{A}{t} + \cfrac{B}{{t - 1}}$

$\Rightarrow$ $1 = A\left( {t - 1} \right) + Bt$ ..(i)
Putting $t = 1$ in $(i)$ ,

we get $B = 1$

Putting $t = 0$ in $(i)$ ,

we get $1 = A\left( {0 - 1} \right)$ $\Rightarrow$ $A = - 1$

$\therefore$ $\cfrac{1}{{t\left( {t - 1} \right)}} = \cfrac{{ - 1}}{t} + \cfrac{1}{{t - 1}}$

$\Rightarrow$ $I = \int {\left( {\cfrac{{ - 1}}{t} + \cfrac{1}{{t - 1}}} \right)} dt = - \log \left| t \right| + \log \left| {t - 1} \right| + C$

$= - \log \left| {{e^x}} \right| + \log \left| {{e^x} - 1} \right| + C = \log \left| {\cfrac{{{e^x} - 1}}{{{e^x}}}} \right| + C$

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