What is the integral of (x^3-5)(x^2+4) ?

The integral of (x^3-5)(x^2+4) is (x^6)/6+x^4-(5x^3)/3-20x+C

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integral of (x^3-5)(x^2+4)

\int (x^{3} - 5) (x^{2} + 4) d x

\frac{x ^{6}}{6} + x^{4} - \frac{5 x ^{3}}{3} - 20 x + C

Solution steps

\int (x^{3} - 5) (x^{2} + 4) d x

Expand

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

= \int x^{5} + 4 x^{3} - 5 x^{2} - 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^{5} d x + \int 4 x^{3} d x - \int 5 x^{2} d x - \int 20 d x

\int x^{5} d x = \frac{x ^{6}}{6}

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

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

\int 20 d x = 20 x

= \frac{x ^{6}}{6} + x^{4} - \frac{5 x ^{3}}{3} - 20 x

Add a constant to the solution

= \frac{x ^{6}}{6} + x^{4} - \frac{5 x ^{3}}{3} - 20 x + C

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What is the integral of (x^3-5)(x^2+4) ?
The integral of (x^3-5)(x^2+4) is (x^6)/6+x^4-(5x^3)/3-20x+C