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

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

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

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

\frac{x ^{2}}{2} + 2 x + ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ + C

Solution steps

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

Expand

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

\int \frac{x ^{3}}{x ^{2} - 2 x - 3} d x = \frac{x ^{2}}{2} + 2 x + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{27}{4} ln ∣ x - 3 ∣

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

= \frac{x ^{2}}{2} + 2 x + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{27}{4} ln ∣ x - 3 ∣ + 3 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣)

Simplify

\frac{x ^{2}}{2} + 2 x + \frac{1}{4} ln ∣ x + 1 ∣ + \frac{27}{4} ln ∣ x - 3 ∣ + 3 (\frac{1}{4} ln ∣ x + 1 ∣ + \frac{3}{4} ln ∣ x - 3 ∣) : \frac{x ^{2}}{2} + 2 x + ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣

= \frac{x ^{2}}{2} + 2 x + ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x + ln ∣ x + 1 ∣ + 9 ln ∣ x - 3 ∣ + C

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