What is the integral of (x-4)(x+1)(x+5) ?

The integral of (x-4)(x+1)(x+5) is (x^4)/4+(2x^3)/3-(19x^2)/2-20x+C

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integral of (x-4)(x+1)(x+5)

\int (x - 4) (x + 1) (x + 5) d x

\frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{19 x ^{2}}{2} - 20 x + C

Solution steps

\int (x - 4) (x + 1) (x + 5) d x

Expand

(x - 4) (x + 1) (x + 5) : x^{3} + 2 x^{2} - 19 x - 20

= \int x^{3} + 2 x^{2} - 19 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^{3} d x + \int 2 x^{2} d x - \int 19 x d x - \int 20 d x

\int x^{3} d x = \frac{x ^{4}}{4}

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

\int 19 x d x = \frac{19 x ^{2}}{2}

\int 20 d x = 20 x

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{19 x ^{2}}{2} - 20 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{2 x ^{3}}{3} - \frac{19 x ^{2}}{2} - 20 x + C

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What is the integral of (x-4)(x+1)(x+5) ?
The integral of (x-4)(x+1)(x+5) is (x^4)/4+(2x^3)/3-(19x^2)/2-20x+C