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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

\int \frac{6}{x ^{3} - 2 x ^{2} - 3 x} d x = 6 (- \frac{1}{3} ln ∣ x ∣ + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{12} ln ∣ x - 3 ∣)

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

Simplify

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

= - ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣

Add a constant to the solution

= - ln ∣ x + 1 ∣ - ln ∣ x - 3 ∣ + 2 ln ∣ x ∣ + C

x \to 1 + lim (\frac{4}{x - 1})

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

x \to 0 lim (x^{2 x})

\frac{d}{d v} (v i)

n = 1 \sum \infty 2 (0.1)^{n}

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