What is the integral of (2x+1)/(x^3-3x^2-4x) ?

The integral of (2x+1)/(x^3-3x^2-4x) is -1/5 ln|x+1|+9/20 ln|x-4|-1/4 ln|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+1)/(x^3-3x^2-4x)

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

- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{9}{20} ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + C

Solution steps

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

Expand

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

\int \frac{2 x}{x ^{3} - 3 x ^{2} - 4 x} d x = 2 (- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣)

\int \frac{1}{x ^{3} - 3 x ^{2} - 4 x} d x = - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{20} ln ∣ x - 4 ∣

= 2 (- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{20} ln ∣ x - 4 ∣

Simplify

2 (- \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{5} ln ∣ x - 4 ∣) - \frac{1}{4} ln ∣ x ∣ + \frac{1}{5} ln ∣ x + 1 ∣ + \frac{1}{20} ln ∣ x - 4 ∣ : - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{9}{20} ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{9}{20} ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{5} ln ∣ x + 1 ∣ + \frac{9}{20} ln ∣ x - 4 ∣ - \frac{1}{4} ln ∣ x ∣ + C

\int 3 x x + 1 d x

\int 5 x (9 - x^{- 9}) d x

\int cos (2 x + 4) d x

\int sec (x) (7 sec (x) + 4 tan (x)) d x

\int \frac{x}{7 + x ^{2}} d x

What is the integral of (2x+1)/(x^3-3x^2-4x) ?
The integral of (2x+1)/(x^3-3x^2-4x) is -1/5 ln|x+1|+9/20 ln|x-4|-1/4 ln|x|+C