Integrate the function: $\frac{e^{x}}{(1+e^{x})(2+e^{x})}$

(N/A) Let $I = \int \frac{e^{x}}{(1+e^{x})(2+e^{x})} dx$.
Substitute $e^{x} = t$,which implies $e^{x} dx = dt$.
The integral becomes $\int \frac{dt}{(t+1)(t+2)}$.
Using partial fractions,we write $\frac{1}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$.
Solving for constants,$1 = A(t+2) + B(t+1)$.
For $t = -1$,$A = 1$. For $t = -2$,$B = -1$.
Thus,$\int \left( \frac{1}{t+1} - \frac{1}{t+2} \right) dt = \log|t+1| - \log|t+2| + C$.
Using the property $\log a - \log b = \log(\frac{a}{b})$,we get $\log|\frac{t+1}{t+2}| + C$.
Substituting back $t = e^{x}$,the final result is $\log|\frac{1+e^{x}}{2+e^{x}}| + C$.