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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Complete the square

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

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

Apply u-substitution

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

Apply the Power Rule

= - \frac{1}{v}

Substitute back

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

Add a constant to the solution

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

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