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

The integral of (x^2-x-6)/(x+2) is -5x+1/2 (x+2)^2-10+C

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

\int \frac{x ^{2} - x - 6}{x + 2} d x

- 5 x + \frac{1}{2} (x + 2)^{2} - 10 + C

Solution steps

\int \frac{x ^{2} - x - 6}{x + 2} d x

Expand

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

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

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

Simplify

\frac{( x + 2 ) ^{2}}{2} - 4 (x + 2) + 4 ln ∣ x + 2 ∣ - (x + 2 - 2 ln ∣ x + 2 ∣) - 6 ln ∣ x + 2 ∣ : - 5 x + \frac{1}{2} (x + 2)^{2} - 10

= - 5 x + \frac{1}{2} (x + 2)^{2} - 10

Add a constant to the solution

= - 5 x + \frac{1}{2} (x + 2)^{2} - 10 + C

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What is the integral of (x^2-x-6)/(x+2) ?
The integral of (x^2-x-6)/(x+2) is -5x+1/2 (x+2)^2-10+C