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

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

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

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

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

Solution steps

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

Expand

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

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

y = x^{2} - 1, y = - x + 2, x = 0, x = 1

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

\frac{d}{d x} (9 x sin (x^{2}))

y^{^{''}} + 9 y^{^{'}} = 0

y^{^{'}} = 3 x^{2} y - 8 e^{(x^{3} - 4 x)}

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