What is the integral of (x-8)(x^2-20x+96) ?

The integral of (x-8)(x^2-20x+96) is (x^4)/4-(28x^3)/3+128x^2-768x+C

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integral of (x-8)(x^2-20x+96)

\int (x - 8) (x^{2} - 20 x + 96) d x

\frac{x ^{4}}{4} - \frac{28 x ^{3}}{3} + 128 x^{2} - 768 x + C

Solution steps

\int (x - 8) (x^{2} - 20 x + 96) d x

Expand

(x - 8) (x^{2} - 20 x + 96) : x^{3} - 28 x^{2} + 256 x - 768

= \int x^{3} - 28 x^{2} + 256 x - 768 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 28 x^{2} d x + \int 256 x d x - \int 768 d x

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

\int 28 x^{2} d x = \frac{28 x ^{3}}{3}

\int 256 x d x = 128 x^{2}

\int 768 d x = 768 x

= \frac{x ^{4}}{4} - \frac{28 x ^{3}}{3} + 128 x^{2} - 768 x

Add a constant to the solution

= \frac{x ^{4}}{4} - \frac{28 x ^{3}}{3} + 128 x^{2} - 768 x + C

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What is the integral of (x-8)(x^2-20x+96) ?
The integral of (x-8)(x^2-20x+96) is (x^4)/4-(28x^3)/3+128x^2-768x+C