What is the integral of (x-1)(x+2)(x+4) ?

The integral of (x-1)(x+2)(x+4) is (x^4)/4+(5x^3)/3+x^2-8x+C

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integral of (x-1)(x+2)(x+4)

\int (x - 1) (x + 2) (x + 4) d x

\frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} + x^{2} - 8 x + C

Solution steps

\int (x - 1) (x + 2) (x + 4) d x

Expand

(x - 1) (x + 2) (x + 4) : x^{3} + 5 x^{2} + 2 x - 8

= \int x^{3} + 5 x^{2} + 2 x - 8 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 5 x^{2} d x + \int 2 x d x - \int 8 d x

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

\int 5 x^{2} d x = \frac{5 x ^{3}}{3}

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

\int 8 d x = 8 x

= \frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} + x^{2} - 8 x

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{5 x ^{3}}{3} + x^{2} - 8 x + C

What is the integral of (x-1)(x+2)(x+4) ?
The integral of (x-1)(x+2)(x+4) is (x^4)/4+(5x^3)/3+x^2-8x+C