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

\frac{d}{d x} (\frac{x}{1 - ln ( x - 6 )})

\frac{\partial}{\partial y} (\frac{x}{y + z})

\int \frac{3 - x ^{2}}{x} d x

8 csc (x) + 2

f (x) = x - 2 sin (x)

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