What is the integral of 1/(1+e^{-y)} ?

The integral of 1/(1+e^{-y)} is ln|1+e^{-y}|+y+C

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integral of 1/(1+e^{-y)}

\int \frac{1}{1 + e ^{- y}} d y

ln 1 + e^{- y} + y + C

Solution steps

\int \frac{1}{1 + e ^{- y}} d y

\frac{1}{1 + e ^{- y}} = - \frac{e ^{- y}}{1 + e ^{- y}} + 1

= \int - \frac{e ^{- y}}{1 + e ^{- y}} + 1 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{e ^{- y}}{1 + e ^{- y}} d y + \int 1 d y

\int \frac{e ^{- y}}{1 + e ^{- y}} d y = - ln 1 + e^{- y}

\int 1 d y = y

= - (- ln 1 + e^{- y}) + y

Simplify

= ln 1 + e^{- y} + y

Add a constant to the solution

= ln 1 + e^{- y} + y + C

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What is the integral of 1/(1+e^{-y)} ?
The integral of 1/(1+e^{-y)} is ln|1+e^{-y}|+y+C