What is the integral of (x^2-4x-12)/(x+2) ?

The integral of (x^2-4x-12)/(x+2) is -8x+1/2 (x+2)^2-16+C

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integral of (x^2-4x-12)/(x+2)

\int \frac{x ^{2} - 4 x - 12}{x + 2} d x

- 8 x + \frac{1}{2} (x + 2)^{2} - 16 + C

Solution steps

\int \frac{x ^{2} - 4 x - 12}{x + 2} d x

Expand

\frac{x ^{2} - 4 x - 12}{x + 2} : \frac{x ^{2}}{x + 2} - \frac{4 x}{x + 2} - \frac{12}{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 ^{2}}{x + 2} d x - \int \frac{4 x}{x + 2} d x - \int \frac{12}{x + 2} d x

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

\int \frac{4 x}{x + 2} d x = 4 (x + 2 - 2 ln ∣ x + 2 ∣)

\int \frac{12}{x + 2} d x = 12 ln ∣ x + 2 ∣

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

Simplify

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

= - 8 x + \frac{1}{2} (x + 2)^{2} - 16

Add a constant to the solution

= - 8 x + \frac{1}{2} (x + 2)^{2} - 16 + C

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What is the integral of (x^2-4x-12)/(x+2) ?
The integral of (x^2-4x-12)/(x+2) is -8x+1/2 (x+2)^2-16+C