Putting $t={{e}^{x}}$

$dt={{e}^{x}}dx$

$I=\int \frac{dt}{t(t-1)}$

Now putting,

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

$A\left( t-1 \right)+Bt=1$

Now put $t-1=0$

therefore, $t=1$

$A(0)+B(1)=1$

$B=1$

Now put $t=0$

$A(0-1)+B(0)=1$

$A=-1$

Now from equation (1) we get

$\frac{t}{t\left( t-1 \right)}=\frac{-1}{t}+\frac{1}{t-1}$
$\int \frac{1}{t\left( t-1 \right)}dt=\int \frac{1}{t}dt+\int \frac{1}{t-1}dt$

$=\log t+\log \left| t-1 \right|+c$

$=\log \left| \frac{t-1}{t} \right|+c$

$=\log \left| \frac{{{e}^{x}}-1}{{{e}^{x}}} \right|+c$