What is the integral of (x^2-4)/(x^3-12x+9) ?

The integral of (x^2-4)/(x^3-12x+9) is 1/3 ln|x-3|+1/3 ln|4x^2+12x-12|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-4)/(x^3-12x+9)

\int \frac{x ^{2} - 4}{x ^{3} - 12 x + 9} d x

\frac{1}{3} ln ∣ x - 3 ∣ + \frac{1}{3} ln 4 x^{2} + 12 x - 12 + C

Solution steps

\int \frac{x ^{2} - 4}{x ^{3} - 12 x + 9} d x

Take the partial fraction of

\frac{x ^{2} - 4}{x ^{3} - 12 x + 9} : \frac{1}{3 ( x - 3 )} + \frac{2 x + 3}{3 ( x ^{2} + 3 x - 3 )}

= \int \frac{1}{3 ( x - 3 )} + \frac{2 x + 3}{3 ( x ^{2} + 3 x - 3 )} 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

= \int \frac{1}{3 ( x - 3 )} d x + \int \frac{2 x + 3}{3 ( x ^{2} + 3 x - 3 )} d x

\int \frac{1}{3 ( x - 3 )} d x = \frac{1}{3} ln ∣ x - 3 ∣

\int \frac{2 x + 3}{3 ( x ^{2} + 3 x - 3 )} d x = \frac{1}{3} ln 4 x^{2} + 12 x - 12

= \frac{1}{3} ln ∣ x - 3 ∣ + \frac{1}{3} ln 4 x^{2} + 12 x - 12

Add a constant to the solution

= \frac{1}{3} ln ∣ x - 3 ∣ + \frac{1}{3} ln 4 x^{2} + 12 x - 12 + C

What is the integral of (x^2-4)/(x^3-12x+9) ?
The integral of (x^2-4)/(x^3-12x+9) is 1/3 ln|x-3|+1/3 ln|4x^2+12x-12|+C