What is the integral of (x+4)(x-5) ?

The integral of (x+4)(x-5) is (x^3)/3-(x^2)/2-20x+C

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integral of (x+4)(x-5)

\int (x + 4) (x - 5) d x

\frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 20 x + C

Solution steps

\int (x + 4) (x - 5) d x

Expand

(x + 4) (x - 5) : x^{2} - x - 20

= \int x^{2} - x - 20 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 20 d x

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

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

\int 20 d x = 20 x

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 20 x

Add a constant to the solution

= \frac{x ^{3}}{3} - \frac{x ^{2}}{2} - 20 x + C

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

\frac{d}{d x} (\frac{x - 1}{x + 1})

(x^{2} - 7 x + 8) 25 - x^{2}

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

\frac{\partial}{\partial y} (\frac{3 x + 1}{x + 2})

What is the integral of (x+4)(x-5) ?
The integral of (x+4)(x-5) is (x^3)/3-(x^2)/2-20x+C