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

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

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

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

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

Solution steps

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

Factor

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Simplify

e^{x} \frac{1}{e ^{2 x} + 4} : \frac{e ^{x}}{e ^{2 x} + 4}

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

Apply u-substitution

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

= \int \frac{1}{u ^{2} + 4} 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 = arctan (v)

= \frac{1}{2} arctan (v)

Substitute back

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

Add a constant to the solution

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

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