What is the integral of (e^{-x})/(4-e^{-2x)} ?

The integral of (e^{-x})/(4-e^{-2x)} is -1/4 (ln|(e^{-x})/2+1|-ln|(e^{-x})/2-1|)+C

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integral of (e^{-x})/(4-e^{-2x)}

\int \frac{e ^{- x}}{4 - e ^{- 2 x}} d x

- \frac{1}{4} (ln \frac{e ^{- x}}{2} + 1 - ln \frac{e ^{- x}}{2} - 1) + C

Solution steps

\int \frac{e ^{- x}}{4 - e ^{- 2 x}} d x

Apply u-substitution

: \int - \frac{1}{4 - u ^{2}} d u

= \int - \frac{1}{4 - u ^{2}} d u

Take the constant out:

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

= - \int \frac{1}{4 - u ^{2}} d u

Apply Integral Substitution

= - \int \frac{1}{2 ( - v ^{2} + 1 )} d v

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}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

- \frac{1}{2} \frac{ln \frac{e ^{- x}}{2} + 1}{2} - \frac{ln \frac{e ^{- x}}{2} - 1}{2} : - \frac{1}{4} (ln \frac{e ^{- x}}{2} + 1 - ln \frac{e ^{- x}}{2} - 1)

= - \frac{1}{4} (ln \frac{e ^{- x}}{2} + 1 - ln \frac{e ^{- x}}{2} - 1)

Add a constant to the solution

= - \frac{1}{4} (ln \frac{e ^{- x}}{2} + 1 - ln \frac{e ^{- x}}{2} - 1) + C

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What is the integral of (e^{-x})/(4-e^{-2x)} ?
The integral of (e^{-x})/(4-e^{-2x)} is -1/4 (ln|(e^{-x})/2+1|-ln|(e^{-x})/2-1|)+C