What is the integral of (e^{3x})/(1-e^{3x)} ?

The integral of (e^{3x})/(1-e^{3x)} is -1/3 ln|1-e^{3x}|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^{3x})/(1-e^{3x)}

\int \frac{e ^{3 x}}{1 - e ^{3 x}} d x

- \frac{1}{3} ln 1 - e^{3 x} + C

Solution steps

\int \frac{e ^{3 x}}{1 - e ^{3 x}} d x

Apply u-substitution

= \int - \frac{1}{3 u} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 - e^{3 x}

= - \frac{1}{3} ln 1 - e^{3 x}

Add a constant to the solution

= - \frac{1}{3} ln 1 - e^{3 x} + C

\int \frac{x ^{2} + 1}{x ^{2} - 4} d x

\int 400 e^{- 0.1 x} + 8 d x

\int x^{4} e (1 - x^{5}) d x

\int \frac{2}{4 - 9 x ^{2}} d x

\int (2^{x} - 4 sin (x)) d x

What is the integral of (e^{3x})/(1-e^{3x)} ?
The integral of (e^{3x})/(1-e^{3x)} is -1/3 ln|1-e^{3x}|+C