$\int\frac{e^xdx}{(2+e^x)(e^x+1)}$ is:

$(e^x+2)(e^x+1)+C$ $\log(\frac{e^x+1}{e^x+2})+C$ $\frac{1}{2}\log(\frac{e^x+1}{e^x+2})+C$ $\log(\frac{e^x+2}{e^x+1})+C$

$\log(\frac{e^x+1}{e^x+2})+C$

$I=\int\frac{e^xdx}{(2+e^x)(e^x+1)}$. Put $e^x = t$ to get: $e^x dx = dt$ $⇒I=\int\frac{dt}{(t+2)(t+1)}=\int\frac{[(t+2)-(t+1)]dt}{(t+2)(t+1)}=\int(\frac{1}{t+1}-\frac{1}{t+2})dt$ $=ln(t+1)-ln(t+2)+c=ln(\frac{t+1}{t+2})+c=ln(\frac{e^x+1}{e^x+2})+c$