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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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