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

The integral of (x+6)(x-3) is (x^3)/3+(3x^2)/2-18x+C

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

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

\frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} - 18 x + C

Solution steps

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

Expand

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

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

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

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

\int 18 d x = 18 x

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} - 18 x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{3 x ^{2}}{2} - 18 x + C

\frac{\partial}{\partial x} (x \cdot x^{2} + y^{2})

x \to 0 lim ((1 - 5 x)^{\frac{2}{x}})

\frac{\partial}{\partial y} (x^{2} y^{2} e^{y^{2}})

n = 0 \sum \infty (- \frac{1}{5})^{n}

\frac{d}{d x} (\frac{1}{x \frac{3}{2}})

What is the integral of (x+6)(x-3) ?
The integral of (x+6)(x-3) is (x^3)/3+(3x^2)/2-18x+C