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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - e^{2 w}

= - \frac{1}{2} ln 1 - e^{2 w}

Add a constant to the solution

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

\frac{2 x}{9 x ^{2} - 6 x - 8}

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