What is the integral of (e^{-3x})/(2+e^{-3x)} ?

The integral of (e^{-3x})/(2+e^{-3x)} is -1/3 ln|2+e^{-3x}|+C

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integral of (e^{-3x})/(2+e^{-3x)}

\int \frac{e ^{- 3 x}}{2 + e ^{- 3 x}} d x

- \frac{1}{3} ln 2 + e^{- 3 x} + C

Solution steps

\int \frac{e ^{- 3 x}}{2 + e ^{- 3 x}} d x

Apply u-substitution

: \int - \frac{1}{3 u} d u

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 2 + e^{- 3 x}

= - \frac{1}{3} ln 2 + e^{- 3 x}

Add a constant to the solution

= - \frac{1}{3} ln 2 + e^{- 3 x} + C

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What is the integral of (e^{-3x})/(2+e^{-3x)} ?
The integral of (e^{-3x})/(2+e^{-3x)} is -1/3 ln|2+e^{-3x}|+C