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

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

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integral of (e^{-2x})/(1+e^{-x)}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= - (u - ln ∣ u ∣)

Substitute back

u = 1 + e^{- x}

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 x^{2} - 2 x y + x - 3 y)

n = 0 \sum \infty \frac{1}{n ^{2} + 3}

x \to 1 lim (- x^{2})

\int 8 sin^{4} (x) cos (x) d x

\frac{d}{d x} (\frac{2}{2 - x})

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