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

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

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

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

e^{- e^{- x}} + C

Solution steps

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

Apply exponent rule:

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

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

= \int e^{- x - e^{- x}} d x

Apply u-substitution

= \int - e^{- u} d u

Take the constant out:

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

= - \int e^{- u} d u

Apply u-substitution

= - \int - e^{v} d v

Take the constant out:

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

= - (- \int e^{v} d v)

Use the common integral:

\int e^{v} d v = e^{v}

= - (- e^{v})

Substitute back

= - (- e^{- e^{- x}})

Simplify

= e^{- e^{- x}}

Add a constant to the solution

= e^{- e^{- x}} + C

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