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

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

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

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

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

Solution steps

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

Expand

\frac{x ^{2} - 2}{x ^{2} - x - 2} : \frac{x ^{2}}{x ^{2} - x - 2} - \frac{2}{x ^{2} - x - 2}

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

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

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

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

Simplify

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

Add a constant to the solution

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

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