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

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

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

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

- \frac{1}{c} (- ln (\frac{∣ c + e ^{- x} ∣}{∣ c ∣}) + ln \frac{c + e ^{- x}}{c} - 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u ( u - c )} d u

= \int - \frac{1}{u ( u - c )} d u

Take the constant out:

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

= - \int \frac{1}{u ( u - c )} d u

u - c = c (\frac{u}{c} - 1)

= - \int \frac{1}{u c ( \frac{u}{c} - 1 )} d u

Take the constant out:

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

= - \frac{1}{c} \cdot \int \frac{1}{u ( \frac{u}{c} - 1 )} d u

Apply u-substitution

: \int \frac{1}{v ( v - 1 )} d v

= - \frac{1}{c} \cdot \int \frac{1}{v ( v - 1 )} d v

Take the partial fraction of

\frac{1}{v ( v - 1 )} : - \frac{1}{v} + \frac{1}{v - 1}

= - \frac{1}{c} \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

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}{c} (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= - \frac{1}{c} (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Substitute back

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

Apply absolute rule:

\frac{a}{b} = \frac{∣ a ∣}{∣ b ∣}

\frac{c + e ^{- x}}{c} = \frac{∣ c + e ^{- x} ∣}{∣ c ∣}

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

Add a constant to the solution

= - \frac{1}{c} (- ln (\frac{∣ c + e ^{- x} ∣}{∣ c ∣}) + ln \frac{c + e ^{- x}}{c} - 1) + C

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