What is the integral of 1/(x^2-13x+42) ?

The integral of 1/(x^2-13x+42) is -ln|x-6|+ln|x-7|+C

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integral of 1/(x^2-13x+42)

\int \frac{1}{x ^{2} - 13 x + 42} d x

- ln ∣ x - 6 ∣ + ln ∣ x - 7 ∣ + C

Solution steps

\int \frac{1}{x ^{2} - 13 x + 42} d x

Take the partial fraction of

\frac{1}{x ^{2} - 13 x + 42} : - \frac{1}{x - 6} + \frac{1}{x - 7}

= \int - \frac{1}{x - 6} + \frac{1}{x - 7} 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 - 6} d x + \int \frac{1}{x - 7} d x

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

\int \frac{1}{x - 7} d x = ln ∣ x - 7 ∣

= - ln ∣ x - 6 ∣ + ln ∣ x - 7 ∣

Add a constant to the solution

= - ln ∣ x - 6 ∣ + ln ∣ x - 7 ∣ + C

What is the integral of 1/(x^2-13x+42) ?
The integral of 1/(x^2-13x+42) is -ln|x-6|+ln|x-7|+C