What is the integral of (x+2)(x-1)(x-3) ?

The integral of (x+2)(x-1)(x-3) is (x^4)/4-(2x^3)/3-(5x^2)/2+6x+C

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integral of (x+2)(x-1)(x-3)

\int (x + 2) (x - 1) (x - 3) d x

\frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 6 x + C

Solution steps

\int (x + 2) (x - 1) (x - 3) d x

Expand

(x + 2) (x - 1) (x - 3) : x^{3} - 2 x^{2} - 5 x + 6

= \int x^{3} - 2 x^{2} - 5 x + 6 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 5 x d x + \int 6 d x

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

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

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

\int 6 d x = 6 x

= \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 6 x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 6 x + C

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What is the integral of (x+2)(x-1)(x-3) ?
The integral of (x+2)(x-1)(x-3) is (x^4)/4-(2x^3)/3-(5x^2)/2+6x+C