What is the integral of 1/(x^3-x^2-9x+9) ?

The integral of 1/(x^3-x^2-9x+9) is -1/8 ln|x-1|+1/24 ln|x+3|+1/12 ln|x-3|+C

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integral of 1/(x^3-x^2-9x+9)

\int \frac{1}{x ^{3} - x ^{2} - 9 x + 9} d x

- \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{x ^{3} - x ^{2} - 9 x + 9} d x

Take the partial fraction of

\frac{1}{x ^{3} - x ^{2} - 9 x + 9} : - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( x - 3 )}

= \int - \frac{1}{8 ( x - 1 )} + \frac{1}{24 ( x + 3 )} + \frac{1}{12 ( 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{1}{8 ( x - 1 )} d x + \int \frac{1}{24 ( x + 3 )} d x + \int \frac{1}{12 ( x - 3 )} d x

\int \frac{1}{8 ( x - 1 )} d x = \frac{1}{8} ln ∣ x - 1 ∣

\int \frac{1}{24 ( x + 3 )} d x = \frac{1}{24} ln ∣ x + 3 ∣

\int \frac{1}{12 ( x - 3 )} d x = \frac{1}{12} ln ∣ x - 3 ∣

= - \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ x - 1 ∣ + \frac{1}{24} ln ∣ x + 3 ∣ + \frac{1}{12} ln ∣ x - 3 ∣ + C

What is the integral of 1/(x^3-x^2-9x+9) ?
The integral of 1/(x^3-x^2-9x+9) is -1/8 ln|x-1|+1/24 ln|x+3|+1/12 ln|x-3|+C