Question

$\int \frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}dx$

Answer 1

$$\begin{gathered} \int \frac{e^{x}dx}{\left(1+e^x\right)\left(2+e^x\right)} \\ \mathrm{let}{e}^x=t \\ =e^{x}dx=dt \\ =\int \frac{dt}{\left(1+t\right)\left(2+t\right)} \\ =\int \frac{dt}{2+t+2t+t^2} \\ =\int \frac{dt}{t^2+3t+2} \\ =\int \frac{dt}{t^2+3t+{\left({\displaystyle \frac{3}{2}}\right)}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2+2} \\ =\int \frac{dt}{{\left(t+{\displaystyle \frac{3}{2}}\right)}^2-{\displaystyle \frac{9}{4}}+2} \\ =\int \frac{dt}{{\left(t+{\displaystyle \frac{3}{2}}\right)}^2-{\displaystyle \frac{1}{4}}} \\ =\int \frac{dt}{{\left(t+{\displaystyle \frac{3}{2}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2} \end{gathered}$$

$$\begin{gathered} =\frac{1}{2\times {\displaystyle \frac{1}{2}}}\log \left|\frac{t+{\displaystyle \frac{3}{2}}-{\displaystyle \frac{1}{2}}}{t+{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{2}}}\right|+C \\ =\log \left|\frac{t+1}{t+2}\right|+C \\ =\log \left|\frac{e^x+1}{e^x+2}\right|+C \end{gathered}$$