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

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

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

\int (e^{x} + e^{- x})^{- 1} d x

arctan (e^{x}) + C

Solution steps

\int (e^{x} + e^{- x})^{- 1} d x

Apply exponent rule:

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

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

Apply u-substitution

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

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= arctan (u)

Substitute back

u = e^{x}

= arctan (e^{x})

Add a constant to the solution

= arctan (e^{x}) + C

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