What is the integral of (e^y)/(e^y-1) ?

The integral of (e^y)/(e^y-1) is ln|e^y-1|+C

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integral of (e^y)/(e^y-1)

\int \frac{e ^{y}}{e ^{y} - 1} d y

ln ∣ e^{y} - 1 ∣ + C

Solution steps

\int \frac{e ^{y}}{e ^{y} - 1} d y

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = e^{y} - 1

= ln ∣ e^{y} - 1 ∣

Add a constant to the solution

= ln ∣ e^{y} - 1 ∣ + C

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What is the integral of (e^y)/(e^y-1) ?
The integral of (e^y)/(e^y-1) is ln|e^y-1|+C