What is the integral of 1/(1+e^{-5x)} ?

The integral of 1/(1+e^{-5x)} is 1/5 ln|1+e^{-5x}|+x+C

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integral of 1/(1+e^{-5x)}

\int \frac{1}{1 + e ^{- 5 x}} d x

\frac{1}{5} ln 1 + e^{- 5 x} + x + C

Solution steps

\int \frac{1}{1 + e ^{- 5 x}} d x

Apply u-substitution

: \int - \frac{1}{5 ( 1 + e ^{u} )} d u

= \int - \frac{1}{5 ( 1 + e ^{u} )} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{1 + e ^{u}} d u

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

= - \frac{1}{5} \cdot \int - \frac{e ^{u}}{1 + e ^{u}} + 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \frac{1}{5} (- \int \frac{e ^{u}}{1 + e ^{u}} d u + \int 1 d u)

\int \frac{e ^{u}}{1 + e ^{u}} d u = ln ∣ 1 + e^{u} ∣

\int 1 d u = u

= - \frac{1}{5} (- ln ∣ 1 + e^{u} ∣ + u)

Substitute back

u = - 5 x

= - \frac{1}{5} (- ln 1 + e^{- 5 x} - 5 x)

Simplify

- \frac{1}{5} (- ln 1 + e^{- 5 x} - 5 x) : \frac{1}{5} ln 1 + e^{- 5 x} + x

= \frac{1}{5} ln 1 + e^{- 5 x} + x

Add a constant to the solution

= \frac{1}{5} ln 1 + e^{- 5 x} + x + C

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What is the integral of 1/(1+e^{-5x)} ?
The integral of 1/(1+e^{-5x)} is 1/5 ln|1+e^{-5x}|+x+C