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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 + e^{- y}

= - ln 1 + e^{- y}

Add a constant to the solution

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

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