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

Apply u-substitution

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

= \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

Factor

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

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

Take the constant out:

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

= - \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

y^{^{'}} + t y = t y^{2}

y^{^{''}} - 4 y = \frac{4 x}{e ^{2 x}}

n = 0 \sum \infty \frac{1}{n + e ^{n}}

\frac{d}{d x} (\frac{24}{( x + 2 ) ^{3}})

x^{2} + 3 x y + x \frac{d y}{d x} = 0

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