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

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

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

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

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

Solution steps

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

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

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

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

\int 1 d y = y

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

Simplify

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

Add a constant to the solution

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

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