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

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