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

The integral of (3x+3)/(x^3-x^2-2x) is

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

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

- ln \frac{2}{3} + \frac{2}{3} x + ln - \frac{4}{3} + \frac{2}{3} x - \frac{3}{2} ln ∣ x ∣ + ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + C

Solution steps

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

Expand

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

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

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

= - ln \frac{2}{3} + \frac{2}{3} x + ln - \frac{4}{3} + \frac{2}{3} x - \frac{3}{2} ln ∣ x ∣ + ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣

Add a constant to the solution

= - ln \frac{2}{3} + \frac{2}{3} x + ln - \frac{4}{3} + \frac{2}{3} x - \frac{3}{2} ln ∣ x ∣ + ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 2 ∣ + C

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