What is the integral of (x+6)/(2x^3-18x) ?

The integral of (x+6)/(2x^3-18x) is 1/12 (ln|x+3|+3ln|x-3|-4ln|x|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+6)/(2x^3-18x)

\int \frac{x + 6}{2 x ^{3} - 18 x} d x

\frac{1}{12} (ln ∣ x + 3 ∣ + 3 ln ∣ x - 3 ∣ - 4 ln ∣ x ∣) + C

Solution steps

\int \frac{x + 6}{2 x ^{3} - 18 x} d x

Factor

\frac{x + 6}{2 x ^{3} - 18 x}

= \int \frac{x + 6}{2 x ( x + 3 ) ( x - 3 )} d x

Take the constant out:

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

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

Expand

\frac{x + 6}{x ( x + 3 ) ( x - 3 )} : \frac{x}{x ( x + 3 ) ( x - 3 )} + \frac{6}{x ( x + 3 ) ( x - 3 )}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{6}{x ( x + 3 ) ( x - 3 )} d x = 6 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣)

= \frac{1}{2} (- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + 6 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣))

Simplify

\frac{1}{2} (- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x - 3 ∣ + 6 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣)) : \frac{1}{12} (ln ∣ x + 3 ∣ + 3 ln ∣ x - 3 ∣ - 4 ln ∣ x ∣)

= \frac{1}{12} (ln ∣ x + 3 ∣ + 3 ln ∣ x - 3 ∣ - 4 ln ∣ x ∣)

Add a constant to the solution

= \frac{1}{12} (ln ∣ x + 3 ∣ + 3 ln ∣ x - 3 ∣ - 4 ln ∣ x ∣) + C

\int \frac{3 x + 2}{x ^{2} + 2 x + 5} d x

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

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

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

\int x^{3} 1 + 100 x^{2} d x

What is the integral of (x+6)/(2x^3-18x) ?
The integral of (x+6)/(2x^3-18x) is 1/12 (ln|x+3|+3ln|x-3|-4ln|x|)+C