What is the integral of 1/(x^2-9x+18) ?

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

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integral of 1/(x^2-9x+18)

\int \frac{1}{x ^{2} - 9 x + 18} d x

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

Solution steps

\int \frac{1}{x ^{2} - 9 x + 18} d x

Take the partial fraction of

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

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

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

\int \frac{1}{3 ( x - 6 )} d x = \frac{1}{3} ln ∣ x - 6 ∣

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

Add a constant to the solution

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

\int \frac{25 - x ^{2}}{x ^{2}} d x

y = \frac{x ^{10} - 5 x ^{6} - 12 x}{x ^{6}}

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}, y (1) = 1

\frac{\partial}{\partial x} (r^{2} + s^{2})

\frac{\partial}{\partial x} (e^{3 x} d y)

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