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

Explanation

माना `I=int(e^(x))/((1+e^(x))(2+e^(x)))dx`
`therefore" "I=int(1)/((1+t)(2+t))dt" "{:("माना "e^(x)=t),(rArr" "e^(x)dx=dt):}`
माना `(1)/((1+t)(2+t))=(A)/((1+t))+(B)/((2+t))`
`rArr" "1=A(2+t)+B(1+t)`
`t=-1 तो 1=A(2-1)+0=A=1`
`t=-2 तो 1=0+B(1-2)" "rArr" "B=-1`
`therefore" "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`