What is the integral of (x+3)(x^2-15x-54) ?

The integral of (x+3)(x^2-15x-54) is (x^4)/4-4x^3-(99x^2)/2-162x+C

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integral of (x+3)(x^2-15x-54)

\int (x + 3) (x^{2} - 15 x - 54) d x

\frac{x ^{4}}{4} - 4 x^{3} - \frac{99 x ^{2}}{2} - 162 x + C

Solution steps

\int (x + 3) (x^{2} - 15 x - 54) d x

Expand

(x + 3) (x^{2} - 15 x - 54) : x^{3} - 12 x^{2} - 99 x - 162

= \int x^{3} - 12 x^{2} - 99 x - 162 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 12 x^{2} d x - \int 99 x d x - \int 162 d x

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

\int 12 x^{2} d x = 4 x^{3}

\int 99 x d x = \frac{99 x ^{2}}{2}

\int 162 d x = 162 x

= \frac{x ^{4}}{4} - 4 x^{3} - \frac{99 x ^{2}}{2} - 162 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 4 x^{3} - \frac{99 x ^{2}}{2} - 162 x + C

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What is the integral of (x+3)(x^2-15x-54) ?
The integral of (x+3)(x^2-15x-54) is (x^4)/4-4x^3-(99x^2)/2-162x+C