`int(e^x)/((1+e^x)(2+e^x))dx`

Explanation

`"Let I" = int(e^(x))/((1+e^(x))(2+e^(x)))dx`
`I= int(1)/((1+t)(2+t))dt" "underset(rArr" "e^(x) dx=dt)(" Let " " "e^(x)=t)`
`" Let " (1)/((1+t)(2+t))=(A)/((1+t))+(B)/((2+t))`
`rArr` 1=A(2+t) +B(1+t)
t=-1 then 1=A (2-1) +0 `rArr` A=1
t=- 2 then 1=0 +B (1-2) `rArr` B=-1
`I=int(1)/((1+t))dt -int (1)/((2+t))dt`
`=log |1+t|-log |2+t|+C`
`=log |(1+t)/(2+t)|+C= log |(1+e^(x))/(2+e^(x))|+C`