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

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

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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

Take the constant out:

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

= - \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

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