$\int \frac{e^x}{\left(2+e^x\right)\left(e^x+1\right)} d x=$ (where $C$ is $a$ constant of integration.)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int \frac{e^x}{\left(2+e^xight)\left(e^x+1ight)} d x \ & =\int \frac{d t}{(t+2)(t+1)}=\int\left(\frac{-1}{t+2}+\frac{1}{t+1}ight) d t \ & =-\log |t+2|+\log |t+1|+c \ & =\log \left|\frac{t+1}{t+2}ight|+c \ & =\log \left|\frac{e^x+1}{e^x+2}ight|+c\end{aligned}$

$\int \frac{e^x}{\left(2+e^x\right)\left(e^x+1\right)} d x=$ (where $C$ is $a$ constant of integration.)

Solution