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

The integral of (2x^3-x^2+3)/(x^2-2x-3) is x^2+3x+12ln|x-3|+1+C

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

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

x^{2} + 3 x + 12 ln ∣ x - 3 ∣ + 1 + C

Solution steps

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

Expand

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

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

\int \frac{x ^{2}}{x ^{2} - 2 x - 3} d x = x - 1 - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{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 ∣)

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

Simplify

x^{2} + 4 x + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{27}{2} ln ∣ x - 3 ∣ - (x - 1 - \frac{1}{4} ln ∣ x + 1 ∣ + \frac{9}{4} ln ∣ x - 3 ∣) + 3 (- \frac{1}{4} ln ∣ x + 1 ∣ + \frac{1}{4} ln ∣ x - 3 ∣) : x^{2} + 3 x + 12 ln ∣ x - 3 ∣ + 1

= x^{2} + 3 x + 12 ln ∣ x - 3 ∣ + 1

Add a constant to the solution

= x^{2} + 3 x + 12 ln ∣ x - 3 ∣ + 1 + C

\int (x - 2) (3 x + 1) d x

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What is the integral of (2x^3-x^2+3)/(x^2-2x-3) ?
The integral of (2x^3-x^2+3)/(x^2-2x-3) is x^2+3x+12ln|x-3|+1+C