What is the integral of (x-9)/(x^3-5x^2+6x) ?

The integral of (x-9)/(x^3-5x^2+6x) is -3/2 ln|x|+7/2 ln|x-2|-2ln|x-3|+C

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integral of (x-9)/(x^3-5x^2+6x)

\int \frac{x - 9}{x ^{3} - 5 x ^{2} + 6 x} d x

- \frac{3}{2} ln ∣ x ∣ + \frac{7}{2} ln ∣ x - 2 ∣ - 2 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{x - 9}{x ^{3} - 5 x ^{2} + 6 x} d x

Take the partial fraction of

\frac{x - 9}{x ^{3} - 5 x ^{2} + 6 x} : - \frac{3}{2 x} + \frac{7}{2 ( x - 2 )} - \frac{2}{x - 3}

= \int - \frac{3}{2 x} + \frac{7}{2 ( x - 2 )} - \frac{2}{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{3}{2 x} d x + \int \frac{7}{2 ( x - 2 )} d x - \int \frac{2}{x - 3} d x

\int \frac{3}{2 x} d x = \frac{3}{2} ln ∣ x ∣

\int \frac{7}{2 ( x - 2 )} d x = \frac{7}{2} ln ∣ x - 2 ∣

\int \frac{2}{x - 3} d x = 2 ln ∣ x - 3 ∣

= - \frac{3}{2} ln ∣ x ∣ + \frac{7}{2} ln ∣ x - 2 ∣ - 2 ln ∣ x - 3 ∣

Add a constant to the solution

= - \frac{3}{2} ln ∣ x ∣ + \frac{7}{2} ln ∣ x - 2 ∣ - 2 ln ∣ x - 3 ∣ + C

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What is the integral of (x-9)/(x^3-5x^2+6x) ?
The integral of (x-9)/(x^3-5x^2+6x) is -3/2 ln|x|+7/2 ln|x-2|-2ln|x-3|+C