What is the integral of (8x^2+3x-6)/((x^2-4)(x+2)) ?

The integral of (8x^2+3x-6)/((x^2-4)(x+2)) is 6ln|x+2|+5/(x+2)+2ln|x-2|+C

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integral of (8x^2+3x-6)/((x^2-4)(x+2))

\int \frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} d x

Take the partial fraction of

\frac{8 x ^{2} + 3 x - 6}{( x ^{2} - 4 ) ( x + 2 )} : \frac{6}{x + 2} - \frac{5}{( x + 2 ) ^{2}} + \frac{2}{x - 2}

= \int \frac{6}{x + 2} - \frac{5}{( x + 2 ) ^{2}} + \frac{2}{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{6}{x + 2} d x - \int \frac{5}{( x + 2 ) ^{2}} d x + \int \frac{2}{x - 2} d x

\int \frac{6}{x + 2} d x = 6 ln ∣ x + 2 ∣

\int \frac{5}{( x + 2 ) ^{2}} d x = - \frac{5}{x + 2}

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

= 6 ln ∣ x + 2 ∣ - (- \frac{5}{x + 2}) + 2 ln ∣ x - 2 ∣

Simplify

= 6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣

Add a constant to the solution

= 6 ln ∣ x + 2 ∣ + \frac{5}{x + 2} + 2 ln ∣ x - 2 ∣ + C

\frac{2}{x} - 3 e^{x} + 3 e^{4}

75 (1 - \frac{5}{x})

n = 1 \sum \infty (\frac{2}{5})^{n - 1}

x \to 0 lim (\frac{5 - 5 cos ( x )}{x})

t \to \infty lim (t^{4})

What is the integral of (8x^2+3x-6)/((x^2-4)(x+2)) ?
The integral of (8x^2+3x-6)/((x^2-4)(x+2)) is 6ln|x+2|+5/(x+2)+2ln|x-2|+C