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

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

Popular >

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

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

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

Solution steps

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

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

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

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 ^{- x}}{1 + e ^{- x}} d x + \int 1 d x

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

\int 1 d x = x

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

Simplify

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

Add a constant to the solution

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

x \to + 2 - lim (\frac{x}{x ^{2} + 4})

\frac{d}{d x} ((- x + \frac{π}{2}) tan (x))

\frac{\partial}{\partial y} (x z t^{2} y + 1)

f (x) = 9 - 2 x^{2}

f (x) = 6 x e^{x}, at x = 0

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