What is the integral of (x+4)(x-3) ?

The integral of (x+4)(x-3) is (x^3)/3+(x^2)/2-12x+C

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integral of (x+4)(x-3)

\int (x + 4) (x - 3) d x

\frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x + C

Solution steps

\int (x + 4) (x - 3) d x

Expand

(x + 4) (x - 3) : x^{2} + x - 12

= \int x^{2} + x - 12 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 12 d x

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

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

\int 12 d x = 12 x

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x

Add a constant to the solution

= \frac{x ^{3}}{3} + \frac{x ^{2}}{2} - 12 x + C

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What is the integral of (x+4)(x-3) ?
The integral of (x+4)(x-3) is (x^3)/3+(x^2)/2-12x+C