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

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

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

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

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

Solution steps

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

Expand

\frac{x + 3}{x ^{2} - 2 x - 3} : \frac{x}{x ^{2} - 2 x - 3} + \frac{3}{x ^{2} - 2 x - 3}

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

\int \frac{x}{x ^{2} - 2 x - 3} d x = \frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣

\int \frac{3}{x ^{2} - 2 x - 3} d x = 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

= \frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣ + 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣)

Simplify

\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣ + 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{3}{2} ln ∣ x - 3 ∣

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

Add a constant to the solution

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

\int f (x + b) d x

x + 2

2 x^{3} - 4 x^{2}, at x = 1

\int 14 ln (3 x) d x

y = \frac{7}{x} + 6 sec (x)

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