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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{x ^{3}}{x ^{2} + x - 2} d x = \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣

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

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

= \frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ - (ln ∣ x - 1 ∣ + 2 ln ∣ x + 2 ∣) - ln \frac{4}{3} + \frac{2}{3} x + ln - \frac{2}{3} + \frac{2}{3} x

Simplify

\frac{x ^{2}}{2} - x + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{8}{3} ln ∣ x + 2 ∣ - (ln ∣ x - 1 ∣ + 2 ln ∣ x + 2 ∣) - ln \frac{4}{3} + \frac{2}{3} x + ln - \frac{2}{3} + \frac{2}{3} x : \frac{x ^{2}}{2} - x - \frac{2}{3} ln ∣ x - 1 ∣ + \frac{2}{3} ln ∣ x + 2 ∣ - ln \frac{2}{3} x + \frac{4}{3} + ln \frac{2}{3} x - \frac{2}{3}

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

Add a constant to the solution

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

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