What is the integral of (x^3+1)/(x^2+x-2) ?

The integral of (x^3+1)/(x^2+x-2) is

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integral of (x^3+1)/(x^2+x-2)

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

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

Solution steps

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

Expand

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

Add a constant to the solution

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

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What is the integral of (x^3+1)/(x^2+x-2) ?
The integral of (x^3+1)/(x^2+x-2) is