What is the integral of 1/(2+e^x+e^{-x)} ?

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

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integral of 1/(2+e^x+e^{-x)}

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

- \frac{1}{e ^{x} + 1} + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

2 u + u^{2} + 1 : (u + 1)^{2}

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

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

= - \frac{1}{e ^{x} + 1}

Add a constant to the solution

= - \frac{1}{e ^{x} + 1} + C

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