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

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

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integral of 1/(e^x-1)

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

ln ∣ e^{x} - 1 ∣ - x + C

Solution steps

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

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

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

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 ^{x}}{e ^{x} - 1} d x - \int 1 d x

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

\int 1 d x = x

= ln ∣ e^{x} - 1 ∣ - x

Add a constant to the solution

= ln ∣ e^{x} - 1 ∣ - x + C

\int sin (\frac{x}{2}) d x

f (x) = \frac{1}{cos ( x )}

\frac{d}{d y} (2 x y^{4} e^{y} + 2 x y^{3} + y)

y^{^{''}} + 4 y^{^{'}} + 6 y = cos (x), y (0) = 0, y^{^{'}} (0) = 2

n = 1 \sum \infty \frac{x ^{n}}{2 n - 1}