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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u} d u

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

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}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{- 2 x} + 1

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

Add a constant to the solution

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

x \to - 2 lim (- x^{2} + 2 x - 6)

\int \frac{1}{x ^{3}} sin (\frac{1}{x ^{2}}) d x

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

\frac{d y}{d x} x^{2} y = y - 7

\int_{- 1}^{3} x^{2} - 2 x - 3 d x

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