What is the integral of (2x^2+3x+6)/((x+3)(x^2-4)) ?

The integral of (2x^2+3x+6)/((x+3)(x^2-4)) is 3ln|x+3|-2ln|x+2|+ln|x-2|+C

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integral of (2x^2+3x+6)/((x+3)(x^2-4))

\int \frac{2 x ^{2} + 3 x + 6}{( x + 3 ) ( x ^{2} - 4 )} d x

3 ln ∣ x + 3 ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{2 x ^{2} + 3 x + 6}{( x + 3 ) ( x ^{2} - 4 )} d x

Take the partial fraction of

\frac{2 x ^{2} + 3 x + 6}{( x + 3 ) ( x ^{2} - 4 )} : \frac{3}{x + 3} - \frac{2}{x + 2} + \frac{1}{x - 2}

= \int \frac{3}{x + 3} - \frac{2}{x + 2} + \frac{1}{x - 2} 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{2}{x + 2} d x + \int \frac{1}{x - 2} d x

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

\int \frac{2}{x + 2} d x = 2 ln ∣ x + 2 ∣

\int \frac{1}{x - 2} d x = ln ∣ x - 2 ∣

= 3 ln ∣ x + 3 ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣

Add a constant to the solution

= 3 ln ∣ x + 3 ∣ - 2 ln ∣ x + 2 ∣ + ln ∣ x - 2 ∣ + C

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What is the integral of (2x^2+3x+6)/((x+3)(x^2-4)) ?
The integral of (2x^2+3x+6)/((x+3)(x^2-4)) is 3ln|x+3|-2ln|x+2|+ln|x-2|+C