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

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

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

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

- \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} x - ln - \frac{2}{3} + \frac{2}{3} x) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{1}{x ^{3} + x ^{2} - 2 x} d x = - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣

= - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} x - ln - \frac{2}{3} + \frac{2}{3} x) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣

Add a constant to the solution

= - \frac{1}{3} (ln \frac{4}{3} + \frac{2}{3} x - ln - \frac{2}{3} + \frac{2}{3} x) - \frac{1}{2} ln ∣ x ∣ + \frac{1}{3} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 2 ∣ + C

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