What is the integral of (x^2-x+24)/(x^3+6x) ?

The integral of (x^2-x+24)/(x^3+6x) is 4ln|x|-3/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^2-x+24)/(x^3+6x)

\int \frac{x ^{2} - x + 24}{x ^{3} + 6 x} d x

4 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6}) + C

Solution steps

\int \frac{x ^{2} - x + 24}{x ^{3} + 6 x} d x

Take the partial fraction of

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

= \int \frac{4}{x} + \frac{- 3 x - 1}{x ^{2} + 6} 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{4}{x} d x + \int \frac{- 3 x - 1}{x ^{2} + 6} d x

\int \frac{4}{x} d x = 4 ln ∣ x ∣

\int \frac{- 3 x - 1}{x ^{2} + 6} d x = - \frac{3}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6})

= 4 ln ∣ x ∣ - \frac{3}{2} ln x^{2} + 6 - \frac{1}{6} arctan (\frac{x}{6})

Add a constant to the solution

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

t an g e n t y = x - 3 x^{2}, (3, - 24)

\frac{\partial}{\partial x} (3 x^{2} y^{4} - 2)

\int (5 x - 7) e^{5 x - 7} d x

\frac{d}{d x} (\frac{2 + 3 x}{x ^{2}})

\int sin^{18} (x) cos^{3} (x) d x

What is the integral of (x^2-x+24)/(x^3+6x) ?
The integral of (x^2-x+24)/(x^3+6x) is 4ln|x|-3/2 ln|x^2+6|-1/(sqrt(6))arctan(x/(sqrt(6)))+C