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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{e ^{u}}{2 ( e ^{2 u} + 4 )} d u

= \int - \frac{e ^{u}}{2 ( e ^{2 u} + 4 )} 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{e ^{u}}{e ^{2 u} + 4} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 4} d v

= - \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 4} d v

Apply Integral Substitution

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

Take the constant out:

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

= - \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= - \frac{1}{2} \cdot \frac{1}{2} arctan (w)

Substitute back

= - \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{e ^{- 2 x}}{2})

Simplify

- \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{e ^{- 2 x}}{2}) : - \frac{1}{4} arctan (\frac{1}{2} e^{- 2 x})

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

Add a constant to the solution

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

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