What is the integral of 1/(x^2-5x+6) ?

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

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integral of 1/(x^2-5x+6)

\int \frac{1}{x ^{2} - 5 x + 6} d x

- ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 5 x + 6} d x

Take the partial fraction of

\frac{1}{x ^{2} - 5 x + 6} : - \frac{1}{x - 2} + \frac{1}{x - 3}

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

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

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

= - ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣

Add a constant to the solution

= - ln ∣ x - 2 ∣ + ln ∣ x - 3 ∣ + C

x \to - 2 lim (\frac{x + 2}{x - 2})

x \to \infty lim (\frac{7}{x})

\frac{3 x + 1}{x ^{2} + 1}

\int \frac{272 x ^{2} + 4}{x ^{2} ( x - 4 )} d x

(2 x y - 3 x^{2}) d x + (x^{2} - 2 y) d y = 0

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