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

The integral of ((x-9))/(x^2-x-2) is 10/3 ln|x+1|-7/3 ln|x-2|+C

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integral of ((x-9))/(x^2-x-2)

\int \frac{( x - 9 )}{x ^{2} - x - 2} d x

\frac{10}{3} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{x - 9}{x ^{2} - x - 2} d x

Take the partial fraction of

\frac{x - 9}{x ^{2} - x - 2} : \frac{10}{3 ( x + 1 )} - \frac{7}{3 ( x - 2 )}

= \int \frac{10}{3 ( x + 1 )} - \frac{7}{3 ( x - 2 )} 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{10}{3 ( x + 1 )} d x - \int \frac{7}{3 ( x - 2 )} d x

\int \frac{10}{3 ( x + 1 )} d x = \frac{10}{3} ln ∣ x + 1 ∣

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

= \frac{10}{3} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣

Add a constant to the solution

= \frac{10}{3} ln ∣ x + 1 ∣ - \frac{7}{3} ln ∣ x - 2 ∣ + C

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What is the integral of ((x-9))/(x^2-x-2) ?
The integral of ((x-9))/(x^2-x-2) is 10/3 ln|x+1|-7/3 ln|x-2|+C