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

The integral of (x+2)/(x-2)(x^2-4x) is 2x^2-12x+1/3 (x-2)^3-16ln|x-2|+16+C

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

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

2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16 + C

Solution steps

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

\frac{x + 2}{x - 2} = \frac{4}{x - 2} + 1

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

Expand

(\frac{4}{x - 2} + 1) (x^{2} - 4 x) : \frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2}

= \int \frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2} d x

Expand

\frac{x ^{3} - 2 x ^{2} - 8 x}{x - 2} : \frac{x ^{3}}{x - 2} - \frac{2 x ^{2}}{x - 2} - \frac{8 x}{x - 2}

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} d x - \int \frac{2 x ^{2}}{x - 2} d x - \int \frac{8 x}{x - 2} d x

\int \frac{x ^{3}}{x - 2} d x = 3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12

\int \frac{2 x ^{2}}{x - 2} d x = x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12

\int \frac{8 x}{x - 2} d x = 8 (x - 2 + 2 ln ∣ x - 2 ∣)

= 3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12 - (x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12) - 8 (x - 2 + 2 ln ∣ x - 2 ∣)

Simplify

3 x^{2} + 8 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 12 - (x^{2} + 4 x + 8 ln ∣ x - 2 ∣ - 12) - 8 (x - 2 + 2 ln ∣ x - 2 ∣) : 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16

= 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16

Add a constant to the solution

= 2 x^{2} - 12 x + \frac{1}{3} (x - 2)^{3} - 16 ln ∣ x - 2 ∣ + 16 + C

What is the integral of (x+2)/(x-2)(x^2-4x) ?
The integral of (x+2)/(x-2)(x^2-4x) is 2x^2-12x+1/3 (x-2)^3-16ln|x-2|+16+C