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

x \to 10 lim (3)

x \to 5 lim (x^{2} - 5 x)

x \to \frac{π}{2} lim ((tan (x))^{\frac{x}{2} - x})

(3 x^{2}) (x^{\frac{1}{2}})

\frac{\partial}{\partial x} (\frac{2 x - y}{2})

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