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)

integral

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

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

sec^{22} (x)

\frac{x ^{2}}{49 - x ^{2}}

\frac{y}{x ^{2}}

\frac{3 x ^{2}}{cos ^{2} ( x ^{3} )}

e^{x - 2}

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