What is the integral of 1/(x^2-3x-18) ?

The integral of 1/(x^2-3x-18) is -1/9 ln|x+3|+1/9 ln|x-6|+C

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integral of 1/(x^2-3x-18)

\int \frac{1}{x ^{2} - 3 x - 18} d x

- \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{9} ln ∣ x - 6 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 3 x - 18} d x

Take the partial fraction of

\frac{1}{x ^{2} - 3 x - 18} : - \frac{1}{9 ( x + 3 )} + \frac{1}{9 ( x - 6 )}

= \int - \frac{1}{9 ( x + 3 )} + \frac{1}{9 ( x - 6 )} 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{1}{9 ( x + 3 )} d x + \int \frac{1}{9 ( x - 6 )} d x

\int \frac{1}{9 ( x + 3 )} d x = \frac{1}{9} ln ∣ x + 3 ∣

\int \frac{1}{9 ( x - 6 )} d x = \frac{1}{9} ln ∣ x - 6 ∣

= - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{9} ln ∣ x - 6 ∣

Add a constant to the solution

= - \frac{1}{9} ln ∣ x + 3 ∣ + \frac{1}{9} ln ∣ x - 6 ∣ + C

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What is the integral of 1/(x^2-3x-18) ?
The integral of 1/(x^2-3x-18) is -1/9 ln|x+3|+1/9 ln|x-6|+C