What is the integral of (x+4)/(2x^3-18x) ?

The integral of (x+4)/(2x^3-18x) is 1/36 (ln|x+3|-8ln|x|+7ln|x-3|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x+4)/(2x^3-18x)

\int \frac{x + 4}{2 x ^{3} - 18 x} d x

\frac{1}{36} (ln ∣ x + 3 ∣ - 8 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{x + 4}{2 x ^{3} - 18 x} d x

Factor

\frac{x + 4}{2 x ^{3} - 18 x}

= \int \frac{x + 4}{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 + 4}{x ( x + 3 ) ( x - 3 )} d x

Expand

\frac{x + 4}{x ( x + 3 ) ( x - 3 )} : \frac{x}{x ( x + 3 ) ( x - 3 )} + \frac{4}{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{4}{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{4}{x ( x + 3 ) ( x - 3 )} d x = 4 (\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 ∣ + 4 (\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 ∣ + 4 (\frac{1}{18} ln ∣ x + 3 ∣ - \frac{1}{9} ln ∣ x ∣ + \frac{1}{18} ln ∣ x - 3 ∣)) : \frac{1}{36} (ln ∣ x + 3 ∣ - 8 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣)

= \frac{1}{36} (ln ∣ x + 3 ∣ - 8 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣)

Add a constant to the solution

= \frac{1}{36} (ln ∣ x + 3 ∣ - 8 ln ∣ x ∣ + 7 ln ∣ x - 3 ∣) + C

\int x sin (6 x^{2}) d x

\int \frac{sec ( x )}{1 + tan ( x )} d x

\int 3 x^{10} - x^{5} + 2 x - 9 d x

\int \frac{1 + cos ( y )}{y + sin ( y )} d y

What is the integral of (x+4)/(2x^3-18x) ?
The integral of (x+4)/(2x^3-18x) is 1/36 (ln|x+3|-8ln|x|+7ln|x-3|)+C