What is the integral of (x+3)(x-2) ?

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

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integral of (x+3)(x-2)

\int (x + 3) (x - 2) d x

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + C

Solution steps

\int (x + 3) (x - 2) d x

Expand

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

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

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

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

\int 6 d x = 6 x

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 6 x + C

\frac{d}{d x} (\frac{1}{x ^{4}})

\frac{1}{4 s ^{2} + 1}

\int cos (\frac{1}{2}) x d x

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 5

\frac{\partial}{\partial x} (6 x - 4 y)

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