What is the integral of (-6x^2+3x+5)/(x^3-x) ?

The integral of (-6x^2+3x+5)/(x^3-x) is ln|x+1|+4ln|x-1|-3ln|x^2-1|-5ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (-6x^2+3x+5)/(x^3-x)

\int \frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} d x

ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣ + C

Solution steps

\int \frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} d x

Expand

\frac{- 6 x ^{2} + 3 x + 5}{x ^{3} - x} : - \frac{6 x ^{2}}{x ^{3} - x} + \frac{3 x}{x ^{3} - x} + \frac{5}{x ^{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{6 x ^{2}}{x ^{3} - x} d x + \int \frac{3 x}{x ^{3} - x} d x + \int \frac{5}{x ^{3} - x} d x

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

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

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

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

Simplify

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

= ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - 3 ln x^{2} - 1 - 5 ln ∣ x ∣ + C

\int \frac{cos ( x )}{( 1 + cos ( x ) )} d x

\int x (2 + x) d x

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

\int \frac{3 x}{4 x + 7} d x

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

What is the integral of (-6x^2+3x+5)/(x^3-x) ?
The integral of (-6x^2+3x+5)/(x^3-x) is ln|x+1|+4ln|x-1|-3ln|x^2-1|-5ln|x|+C