What is the integral from 8 to R of 1/(x^2-9x+18) ?

The integral from 8 to R of 1/(x^2-9x+18) is -1/3 (ln|R-3|-ln(5))+1/3 (ln|R-6|-ln(2))

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integral from 8 to R of 1/(x^2-9x+18)

\int_{8}^{R} \frac{1}{x ^{2} - 9 x + 18} d x

- \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5)) + \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

Solution steps

\int_{8}^{R} \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_{8}^{R} - \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_{8}^{R} \frac{1}{3 ( x - 3 )} d x + \int_{8}^{R} \frac{1}{3 ( x - 6 )} d x

\int_{8}^{R} \frac{1}{3 ( x - 3 )} d x = \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5))

\int_{8}^{R} \frac{1}{3 ( x - 6 )} d x = \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

= - \frac{1}{3} (ln ∣ R - 3 ∣ - ln (5)) + \frac{1}{3} (ln ∣ R - 6 ∣ - ln (2))

What is the integral from 8 to R of 1/(x^2-9x+18) ?
The integral from 8 to R of 1/(x^2-9x+18) is -1/3 (ln|R-3|-ln(5))+1/3 (ln|R-6|-ln(2))