What is the integral of (x^2+24x-4)/(x^3-4x) ?

The integral of (x^2+24x-4)/(x^3-4x) is ln|x|-6ln|x+2|+6ln|x-2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2+24x-4)/(x^3-4x)

\int \frac{x ^{2} + 24 x - 4}{x ^{3} - 4 x} d x

ln ∣ x ∣ - 6 ln ∣ x + 2 ∣ + 6 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x ^{2} + 24 x - 4}{x ^{3} - 4 x} d x

Take the partial fraction of

\frac{x ^{2} + 24 x - 4}{x ^{3} - 4 x} : \frac{1}{x} - \frac{6}{x + 2} + \frac{6}{x - 2}

= \int \frac{1}{x} - \frac{6}{x + 2} + \frac{6}{x - 2} 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}{x} d x - \int \frac{6}{x + 2} d x + \int \frac{6}{x - 2} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x + 2} d x = 6 ln ∣ x + 2 ∣

\int \frac{6}{x - 2} d x = 6 ln ∣ x - 2 ∣

= ln ∣ x ∣ - 6 ln ∣ x + 2 ∣ + 6 ln ∣ x - 2 ∣

Add a constant to the solution

= ln ∣ x ∣ - 6 ln ∣ x + 2 ∣ + 6 ln ∣ x - 2 ∣ + C

\frac{\partial}{\partial x} (\frac{n}{x + 1})

x \to - 5 lim (\frac{5}{x + 3})

\frac{\partial}{\partial x} (- 1)

\int \frac{3 x ^{11}}{x ^{2} - x - 6} d x

\int x \cdot x + 6 d x

What is the integral of (x^2+24x-4)/(x^3-4x) ?
The integral of (x^2+24x-4)/(x^3-4x) is ln|x|-6ln|x+2|+6ln|x-2|+C