What is the integral of (7x-9)/(x^2-4x-21) ?

The integral of (7x-9)/(x^2-4x-21) is 3ln|x+3|+4ln|x-7|+C

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integral of (7x-9)/(x^2-4x-21)

\int \frac{7 x - 9}{x ^{2} - 4 x - 21} d x

3 ln ∣ x + 3 ∣ + 4 ln ∣ x - 7 ∣ + C

Solution steps

\int \frac{7 x - 9}{x ^{2} - 4 x - 21} d x

Take the partial fraction of

\frac{7 x - 9}{x ^{2} - 4 x - 21} : \frac{3}{x + 3} + \frac{4}{x - 7}

= \int \frac{3}{x + 3} + \frac{4}{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{3}{x + 3} d x + \int \frac{4}{x - 7} d x

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

\int \frac{4}{x - 7} d x = 4 ln ∣ x - 7 ∣

= 3 ln ∣ x + 3 ∣ + 4 ln ∣ x - 7 ∣

Add a constant to the solution

= 3 ln ∣ x + 3 ∣ + 4 ln ∣ x - 7 ∣ + C

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What is the integral of (7x-9)/(x^2-4x-21) ?
The integral of (7x-9)/(x^2-4x-21) is 3ln|x+3|+4ln|x-7|+C