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

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

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

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

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

Solution steps

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

= - ln e^{- x} + 1

Add a constant to the solution

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

6 e^{- 2 x} cos (4 x)

f (x) = 5 x^{2} - 2 x - 7

y^{^{'}} = 0.04 y - 2000, y (0) = 25000

e^{3 (t - 2)}

\frac{d}{d x} (\frac{2 x}{3 + x ^{2}})

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