What is the integral of (7x-1)/(x^2-x-6) ?

The integral of (7x-1)/(x^2-x-6) is 3ln|x+2|+4ln|x-3|+C

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integral of (7x-1)/(x^2-x-6)

\int \frac{7 x - 1}{x ^{2} - x - 6} d x

3 ln ∣ x + 2 ∣ + 4 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{7 x - 1}{x ^{2} - x - 6} d x

Take the partial fraction of

\frac{7 x - 1}{x ^{2} - x - 6} : \frac{3}{x + 2} + \frac{4}{x - 3}

= \int \frac{3}{x + 2} + \frac{4}{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{3}{x + 2} d x + \int \frac{4}{x - 3} d x

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{4}{x - 3} d x = 4 ln ∣ x - 3 ∣

= 3 ln ∣ x + 2 ∣ + 4 ln ∣ x - 3 ∣

Add a constant to the solution

= 3 ln ∣ x + 2 ∣ + 4 ln ∣ x - 3 ∣ + C

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What is the integral of (7x-1)/(x^2-x-6) ?
The integral of (7x-1)/(x^2-x-6) is 3ln|x+2|+4ln|x-3|+C