What is the integral of (x^3-6x+5)/(3x^2-3x-18) ?

The integral of (x^3-6x+5)/(3x^2-3x-18) is (x^2)/6+1/3 x-3/5 ln|x+2|+14/15 ln|x-3|+C

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integral of (x^3-6x+5)/(3x^2-3x-18)

\int \frac{x ^{3} - 6 x + 5}{3 x ^{2} - 3 x - 18} d x

\frac{x ^{2}}{6} + \frac{1}{3} x - \frac{3}{5} ln ∣ x + 2 ∣ + \frac{14}{15} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x ^{3} - 6 x + 5}{3 x ^{2} - 3 x - 18} d x

Take the partial fraction of

\frac{x ^{3} - 6 x + 5}{3 x ^{2} - 3 x - 18} : \frac{x}{3} + \frac{1}{3} - \frac{3}{5 ( x + 2 )} + \frac{14}{15 ( x - 3 )}

= \int \frac{x}{3} + \frac{1}{3} - \frac{3}{5 ( x + 2 )} + \frac{14}{15 ( x - 3 )} d 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}{3} d x + \int \frac{1}{3} d x - \int \frac{3}{5 ( x + 2 )} d x + \int \frac{14}{15 ( x - 3 )} d x

\int \frac{x}{3} d x = \frac{x ^{2}}{6}

\int \frac{1}{3} d x = \frac{1}{3} x

\int \frac{3}{5 ( x + 2 )} d x = \frac{3}{5} ln ∣ x + 2 ∣

\int \frac{14}{15 ( x - 3 )} d x = \frac{14}{15} ln ∣ x - 3 ∣

= \frac{x ^{2}}{6} + \frac{1}{3} x - \frac{3}{5} ln ∣ x + 2 ∣ + \frac{14}{15} ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{x ^{2}}{6} + \frac{1}{3} x - \frac{3}{5} ln ∣ x + 2 ∣ + \frac{14}{15} ln ∣ x - 3 ∣ + C

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What is the integral of (x^3-6x+5)/(3x^2-3x-18) ?
The integral of (x^3-6x+5)/(3x^2-3x-18) is (x^2)/6+1/3 x-3/5 ln|x+2|+14/15 ln|x-3|+C