What is the integral of x/(3x^2-4) ?

The integral of x/(3x^2-4) is 1/6 ln|3x^2-4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x/(3x^2-4)

\int \frac{x}{3 x ^{2} - 4} d x

\frac{1}{6} ln 3 x^{2} - 4 + C

Solution steps

\int \frac{x}{3 x ^{2} - 4} d x

Apply u-substitution

= \int \frac{1}{6 u} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{6} ln ∣ u ∣

Substitute back

u = 3 x^{2} - 4

= \frac{1}{6} ln 3 x^{2} - 4

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} - 4 + C

What is the integral of x/(3x^2-4) ?
The integral of x/(3x^2-4) is 1/6 ln|3x^2-4|+C