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

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

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

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

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

Solution steps

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

Factor

e^{- x} - 2 : - (- e^{- x} + 2)

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

Take the constant out:

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

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

Apply u-substitution

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

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

Complete the square

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