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

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

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

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

\frac{x ^{3}}{3} - x^{2} - 2 x + C

Solution steps

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

Simplify

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

= \int x^{2} - 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 x^{2} d x - \int 2 x d x - \int 2 d x

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

\int 2 x d x = x^{2}

\int 2 d x = 2 x

= \frac{x ^{3}}{3} - x^{2} - 2 x

Add a constant to the solution

= \frac{x ^{3}}{3} - x^{2} - 2 x + C

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