What is the integral of ((2e^x-1))/(e^x-1) ?

The integral of ((2e^x-1))/(e^x-1) is ln|e^x-1|+x+C

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integral of ((2e^x-1))/(e^x-1)

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

ln ∣ e^{x} - 1 ∣ + x + C

Solution steps

\int \frac{2 e ^{x} - 1}{e ^{x} - 1} d x

Expand

\frac{2 e ^{x} - 1}{e ^{x} - 1} : \frac{2 e ^{x}}{e ^{x} - 1} - \frac{1}{e ^{x} - 1}

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

\int \frac{2 e ^{x}}{e ^{x} - 1} d x = 2 ln ∣ e^{x} - 1 ∣

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

= 2 ln ∣ e^{x} - 1 ∣ - (ln ∣ e^{x} - 1 ∣ - x)

Simplify

= ln ∣ e^{x} - 1 ∣ + x

Add a constant to the solution

= ln ∣ e^{x} - 1 ∣ + x + C

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What is the integral of ((2e^x-1))/(e^x-1) ?
The integral of ((2e^x-1))/(e^x-1) is ln|e^x-1|+x+C