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

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

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

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

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

Solution steps

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

Expand

(x + 3) (x - 4) : 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

\int_{1}^{2} x^{3} + 2 d x

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

\int \frac{1}{x ^{10}} d x

\frac{1}{( 3 - x ) ^{2}}

\int \frac{1}{x ^{2} + 12 x + 61} d x

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