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

The integral of 1/(4e^{-4x)+e^{4x}} is 1/8 arctan(1/2 e^{4x})+C

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integral of 1/(4e^{-4x)+e^{4x}}

\int \frac{1}{4 e ^{- 4 x} + e ^{4 x}} d x

\frac{1}{8} arctan (\frac{1}{2} e^{4 x}) + C

Solution steps

\int \frac{1}{4 e ^{- 4 x} + e ^{4 x}} d x

Factor

4 e^{- 4 x} + e^{4 x} : e^{- 4 x} (e^{4 x} + 2 e^{2 x} + 2) (e^{4 x} - 2 e^{2 x} + 2)

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

Apply u-substitution

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

Expand

e^{- 2 u} (e^{2 u} + 2 e^{u} + 2) (e^{2 u} - 2 e^{u} + 2) : e^{2 u} + 4 e^{- 2 u}

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

Apply u-substitution

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

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}{v ^{2} + 4} d v

Apply Integral Substitution

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{e ^{2 \cdot 2 x}}{2}) : \frac{1}{8} arctan (\frac{1}{2} e^{4 x})

= \frac{1}{8} arctan (\frac{1}{2} e^{4 x})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{1}{2} e^{4 x}) + C

What is the integral of 1/(4e^{-4x)+e^{4x}} ?
The integral of 1/(4e^{-4x)+e^{4x}} is 1/8 arctan(1/2 e^{4x})+C