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

The integral of 1/(1+e^x+e^{-x)} is -2/(sqrt(3))arctan(1/(sqrt(3))(2e^{-x}+1))+C

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

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

- \frac{2}{3} arctan (\frac{1}{3} (2 e^{- x} + 1)) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Complete the square

u + 1 + u^{2} : (u + \frac{1}{2})^{2} + \frac{3}{4}

= - \int \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Apply u-substitution

: \int \frac{4}{4 v ^{2} + 3} d v

= - \int \frac{4}{4 v ^{2} + 3} d v

Take the constant out:

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

= - 4 \cdot \int \frac{1}{4 v ^{2} + 3} d v

Apply Integral Substitution

= - 4 \cdot \int \frac{1}{2 3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= - 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= - 4 \cdot \frac{1}{2 3} arctan (w)

Substitute back

= - 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (e^{- x} + \frac{1}{2}))

Simplify

- 4 \cdot \frac{1}{2 3} arctan (\frac{2}{3} (e^{- x} + \frac{1}{2})) : - \frac{2}{3} arctan (\frac{1}{3} (2 e^{- x} + 1))

= - \frac{2}{3} arctan (\frac{1}{3} (2 e^{- x} + 1))

Add a constant to the solution

= - \frac{2}{3} arctan (\frac{1}{3} (2 e^{- x} + 1)) + C

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What is the integral of 1/(1+e^x+e^{-x)} ?
The integral of 1/(1+e^x+e^{-x)} is -2/(sqrt(3))arctan(1/(sqrt(3))(2e^{-x}+1))+C