What is the integral of x(x+3)(x+6) ?

The integral of x(x+3)(x+6) is (x^4)/4+3x^3+9x^2+C

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integral of x(x+3)(x+6)

\int x (x + 3) (x + 6) d x

\frac{x ^{4}}{4} + 3 x^{3} + 9 x^{2} + C

Solution steps

\int x (x + 3) (x + 6) d x

Expand

x (x + 3) (x + 6) : x^{3} + 9 x^{2} + 18 x

= \int x^{3} + 9 x^{2} + 18 x 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^{3} d x + \int 9 x^{2} d x + \int 18 x d x

\int x^{3} d x = \frac{x ^{4}}{4}

\int 9 x^{2} d x = 3 x^{3}

\int 18 x d x = 9 x^{2}

= \frac{x ^{4}}{4} + 3 x^{3} + 9 x^{2}

Add a constant to the solution

= \frac{x ^{4}}{4} + 3 x^{3} + 9 x^{2} + C

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What is the integral of x(x+3)(x+6) ?
The integral of x(x+3)(x+6) is (x^4)/4+3x^3+9x^2+C