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

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

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

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

6 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)) + C

Solution steps

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

Expand

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

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

\int \frac{6 x ^{2}}{( x - 2 ) ( x + 4 )} d x = 6 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣)

\int \frac{3 x}{( x - 2 ) ( x + 4 )} d x = 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1))

= 6 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1))

Add a constant to the solution

= 6 (x + 1 - \frac{8}{3} ln ∣ x + 4 ∣ + \frac{2}{3} ln ∣ x - 2 ∣) - 3 (\frac{1}{2} ln x^{2} + 2 x - 8 + \frac{1}{6} (ln \frac{1}{3} (x + 1) + 1 - ln \frac{1}{3} (x + 1) - 1)) + C

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