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

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

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

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

arctan (e^{x}) + C

Solution steps

\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

\frac{1}{n + k e ^{n}}

f (x) = 3 (x)^{3} + cos (x)

\int sin^{2020} (d) x d x

\frac{d}{d x} (- sin (3 x) 3)

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