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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 + e^{- x}

= - ln 1 + e^{- x}

Add a constant to the solution

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

\frac{d}{d x} (4 (sin (x) + x cos (x)))

\frac{d y}{d x} = 8 x e^{y}

\frac{d}{d x} (3^{x + 1})

\int_{4}^{5} (x^{3} - π x^{2}) d x

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- x} (8 cos (x) - 6 sin (x))

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