What is the integral of 2/(e^x+e^{-x)} ?

The integral of 2/(e^x+e^{-x)} is 2arctan(e^x)+C

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integral of 2/(e^x+e^{-x)}

\int \frac{2}{e ^{x} + e ^{- x}} d x

2 arctan (e^{x}) + C

Solution steps

\int \frac{2}{e ^{x} + e ^{- x}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{e ^{x} + e ^{- x}} d x

Apply u-substitution

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

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

Use the common integral:

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

= 2 arctan (u)

Substitute back

u = e^{x}

= 2 arctan (e^{x})

Add a constant to the solution

= 2 arctan (e^{x}) + C

\int 6 cos (8 x) d x

\frac{1}{z}

\int (3 x^{2} - 2 x + 5) d x

f (x) = - \frac{3}{x}

y^{^{''}} - 5 y^{^{'}} + 6 y = cos (x) e^{- x}

What is the integral of 2/(e^x+e^{-x)} ?
The integral of 2/(e^x+e^{-x)} is 2arctan(e^x)+C