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

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

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

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

\frac{x}{2} + \frac{1}{2} ln e^{- x} - 2 + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u ( 2 - u )} d u

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - (\frac{ln ∣ e ^{- x} - 1 + 1 ∣}{2} - \frac{ln ∣ e ^{- x} - 1 - 1 ∣}{2})

Simplify

- (\frac{ln ∣ e ^{- x} - 1 + 1 ∣}{2} - \frac{ln ∣ e ^{- x} - 1 - 1 ∣}{2}) : \frac{x}{2} + \frac{1}{2} ln e^{- x} - 2

= \frac{x}{2} + \frac{1}{2} ln e^{- x} - 2

Add a constant to the solution

= \frac{x}{2} + \frac{1}{2} ln e^{- x} - 2 + C

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