What is the integral of (x+1)/(x^3-x^2-6x) ?

The integral of (x+1)/(x^3-x^2-6x) is -1/10 ln|x+2|+4/15 ln|x-3|-1/6 ln|x|+C

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integral of (x+1)/(x^3-x^2-6x)

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

- \frac{1}{10} ln ∣ x + 2 ∣ + \frac{4}{15} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{x + 1}{x ^{3} - x ^{2} - 6 x} : \frac{x}{x ^{3} - x ^{2} - 6 x} + \frac{1}{x ^{3} - x ^{2} - 6 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} - 6 x} d x + \int \frac{1}{x ^{3} - x ^{2} - 6 x} d x

\int \frac{x}{x ^{3} - x ^{2} - 6 x} d x = - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣

\int \frac{1}{x ^{3} - x ^{2} - 6 x} d x = - \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣

= - \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣

Simplify

- \frac{1}{5} ln ∣ x + 2 ∣ + \frac{1}{5} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣ + \frac{1}{10} ln ∣ x + 2 ∣ + \frac{1}{15} ln ∣ x - 3 ∣ : - \frac{1}{10} ln ∣ x + 2 ∣ + \frac{4}{15} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣

= - \frac{1}{10} ln ∣ x + 2 ∣ + \frac{4}{15} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣

Add a constant to the solution

= - \frac{1}{10} ln ∣ x + 2 ∣ + \frac{4}{15} ln ∣ x - 3 ∣ - \frac{1}{6} ln ∣ x ∣ + C

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What is the integral of (x+1)/(x^3-x^2-6x) ?
The integral of (x+1)/(x^3-x^2-6x) is -1/10 ln|x+2|+4/15 ln|x-3|-1/6 ln|x|+C