ज्ञात करें `int (dx)/(1+e^x)`

Explanation

`[ 1 +e^x =1 +(e^(x/2))^2,]` जो `(a^2+x^2)` के रूप का है, इसलिए `e^(x/2) = tan theta` रखा जा सकता है |
माना की `e^(x/2) = tan theta,` तो `e^(x/2), 1/2 dx = sec^2 theta d theta`
`:." "dx = (2sec^2 theta d theta)/e^(x/2)= (2sec^2 theta)/(tan theta) d theta`
अब, `int = (dx)/(1+e^(x))= int (dx)/(1+(e^(x//2))^2)= int (1/(1+tan^(2)theta))(2 sec^2 theta)/(tan theta)d theta = 2 int cot theta d theta 2 lo g| sin theta| + c" ...(1)"`
`because" "tan theta = e^(x/2)=e^(x/2)/1 :. sin theta = e^(x/2)/sqrt(1+(e^(x//2))^2)=e^(x/2)/sqrt(1+e^x)`
अब (1) से, `int (dx)/(1+e^(x))= 2 log e^(x/2)/sqrt(1+e^(x))+c =log [(e^(x/2))/sqrt(1+e^x)]^2 +c`
`= log (e^x/(1+e^x))+c=x-log (1+e^x)+c`
Second method :
`int (dx)/(1+e^x)=int (1+e^x-e^x)/(1+e^x)dx=int (1e^x/(1+e^x))dx=int dx - int e^x/(1+e^x) dx`