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

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

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

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

- ln ∣ 1 - e^{y} ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 - e^{y}

= - ln ∣ 1 - e^{y} ∣

Add a constant to the solution

= - ln ∣ 1 - e^{y} ∣ + C

x \to 0 lim (\frac{0 x ^{4} + e ^{0 x}}{x ^{4}})

\frac{d}{d x} (x ln^{2} (x))

(y + \frac{3}{y}) (y^{2} - 5)

\frac{d}{d x} (sec^{n - 2} (x))

x \to 0 lim (\frac{1 0 ^{x} - 1}{x})

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