$\cfrac{{{e^x}}}{{\left( {1 + {e^x}} \right)\left( {2 + {e^x}} \right)}}$

Let $I = \int {\cfrac{{{e^x}}}{{\left( {1 + {e^x}} \right)\left( {2 + {e^x}} \right)}}dx}$

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

$\therefore$ $I = \int {\cfrac{{dt}}{{\left( {1 + t} \right)\left( {2 + t} \right)}}}$

Now, we write $\cfrac{1}{{\left( {1 + t} \right)\left( {2 + t} \right)}} = \cfrac{A}{{1 + t}} + \cfrac{B}{{2 + t}}$

$\Rightarrow$ $1 = A\left( {2 + t} \right) + B\left( {1 + t} \right)$

….(i)
Putting $t = - 1$ in (i),

we get $1 = A\left( {2 - 1} \right)$ $\Rightarrow$ $A = 1$
Putting $t = - 2$ in (i),

we get $1 = B\left( {1 - 2} \right)$ $\Rightarrow$ $B = - 1$

$\therefore$ $I = \int {\cfrac{1}{{\left( {1 + t} \right)\left( {2 + t} \right)}}dt} = \int {\left( {\cfrac{1}{{1 + t}} - \cfrac{1}{{2 + t}}} \right)dt}$

$= \log \left( {1 + t} \right) - \log \left( {2 + t} \right) + C = \log \left( {1 + {e^x}} \right) - \log \left( {2 + {e^x}} \right) + C$
$= \log \left( {\cfrac{{1 + {e^x}}}{{2 + {e^x}}}} \right) + C$