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

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

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

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

arctan (e^{z}) + C

Solution steps

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

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^{z}

= arctan (e^{z})

Add a constant to the solution

= arctan (e^{z}) + C

\frac{d}{d x} ((3 + x - 2 x^{2})^{\frac{1}{4}})

\frac{d y}{d x}, 4 x^{2} + 9 y^{2} = 36

y^{^{'}} + 2 x y = x e^{- x^{2}}

3 2690 - 10 x^{2}

\int x sinh (m x) d x

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