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

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

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

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

ln ∣ e^{x} - 1 ∣ + C

Solution steps

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

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^{x} - 1

= ln ∣ e^{x} - 1 ∣

Add a constant to the solution

= ln ∣ e^{x} - 1 ∣ + C

\int_{1}^{2} (x - \frac{1}{x}) d x

\int_{2}^{5} \frac{ln ( x ^{5} )}{x} d x

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

\int 5 e^{2 x} d x

y^{^{''}} + 9 y = t e^{3 t}

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