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

The integral of (e^{-x})/(e^x+e^{-x)} is -1/2 ln|1+e^{-2x}|+C

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

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

- \frac{1}{2} ln 1 + e^{- 2 x} + C

Solution steps

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

Apply u-substitution

= \int - \frac{e ^{2 u}}{1 + e ^{2 u}} d u

Take the constant out:

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

= - \int \frac{e ^{2 u}}{1 + e ^{2 u}} d u

Apply u-substitution

= - \int \frac{1}{2 v} d v

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} ln ∣ v ∣

Substitute back

= - \frac{1}{2} ln 1 + e^{2 (- x)}

Simplify

= - \frac{1}{2} ln 1 + e^{- 2 x}

Add a constant to the solution

= - \frac{1}{2} ln 1 + e^{- 2 x} + C

What is the integral of (e^{-x})/(e^x+e^{-x)} ?
The integral of (e^{-x})/(e^x+e^{-x)} is -1/2 ln|1+e^{-2x}|+C