What is the integral of (10x+6)/(x^2-2x-3) ?

The integral of (10x+6)/(x^2-2x-3) is ln|x+1|+9ln|x-3|+C

integral of (10x+6)/(x^2-2x-3)

\int \frac{10 x + 6}{x ^{2} - 2 x - 3} d x

ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{10 x + 6}{x ^{2} - 2 x - 3} d x

Take the partial fraction of

\frac{10 x + 6}{x ^{2} - 2 x - 3} : \frac{1}{x + 1} + \frac{9}{x - 3}

= \int \frac{1}{x + 1} + \frac{9}{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{1}{x + 1} d x + \int \frac{9}{x - 3} d x

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

\int \frac{9}{x - 3} d x = 9 ln ∣ x - 3 ∣

= ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣

Add a constant to the solution

= ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ + C

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What is the integral of (10x+6)/(x^2-2x-3) ?
The integral of (10x+6)/(x^2-2x-3) is ln|x+1|+9ln|x-3|+C