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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 2/(e^x-1)

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

2 (ln ∣ e^{x} - 1 ∣ - x) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

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

= 2 \cdot \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

= 2 (\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

= 2 (ln ∣ e^{x} - 1 ∣ - x)

Add a constant to the solution

= 2 (ln ∣ e^{x} - 1 ∣ - x) + C

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