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

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

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integral of 1/(e^y-1)

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

ln ∣ e^{y} - 1 ∣ - y + C

Solution steps

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

\frac{1}{e ^{y} - 1} = \frac{e ^{y}}{e ^{y} - 1} - 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{e ^{y}}{e ^{y} - 1} d y = ln ∣ e^{y} - 1 ∣

\int 1 d y = y

= ln ∣ e^{y} - 1 ∣ - y

Add a constant to the solution

= ln ∣ e^{y} - 1 ∣ - y + C

\int 2 x (x^{2} + 3)^{20} d x

\frac{d}{d x} (0.8 x^{2} - 3 x + 5 x^{0.2} - 5200)

\int 4 x - 5 d x

- \frac{3}{( 3 x + 1 ) ^{2}}

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