What is the integral of (5x^2-4)(x+8) ?

The integral of (5x^2-4)(x+8) is (5x^4)/4+(40x^3)/3-2x^2-32x+C

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integral of (5x^2-4)(x+8)

\int (5 x^{2} - 4) (x + 8) d x

\frac{5 x ^{4}}{4} + \frac{40 x ^{3}}{3} - 2 x^{2} - 32 x + C

Solution steps

\int (5 x^{2} - 4) (x + 8) d x

Expand

(5 x^{2} - 4) (x + 8) : 5 x^{3} + 40 x^{2} - 4 x - 32

= \int 5 x^{3} + 40 x^{2} - 4 x - 32 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 5 x^{3} d x + \int 40 x^{2} d x - \int 4 x d x - \int 32 d x

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int 40 x^{2} d x = \frac{40 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 32 d x = 32 x

= \frac{5 x ^{4}}{4} + \frac{40 x ^{3}}{3} - 2 x^{2} - 32 x

Add a constant to the solution

= \frac{5 x ^{4}}{4} + \frac{40 x ^{3}}{3} - 2 x^{2} - 32 x + C

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What is the integral of (5x^2-4)(x+8) ?
The integral of (5x^2-4)(x+8) is (5x^4)/4+(40x^3)/3-2x^2-32x+C