What is the integral of (x^2-6x+8)(x-3) ?

The integral of (x^2-6x+8)(x-3) is (x^4)/4-3x^3+13x^2-24x+C

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integral of (x^2-6x+8)(x-3)

\int (x^{2} - 6 x + 8) (x - 3) d x

\frac{x ^{4}}{4} - 3 x^{3} + 13 x^{2} - 24 x + C

Solution steps

\int (x^{2} - 6 x + 8) (x - 3) d x

Expand

(x^{2} - 6 x + 8) (x - 3) : x^{3} - 9 x^{2} + 26 x - 24

= \int x^{3} - 9 x^{2} + 26 x - 24 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 9 x^{2} d x + \int 26 x d x - \int 24 d x

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

\int 9 x^{2} d x = 3 x^{3}

\int 26 x d x = 13 x^{2}

\int 24 d x = 24 x

= \frac{x ^{4}}{4} - 3 x^{3} + 13 x^{2} - 24 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 3 x^{3} + 13 x^{2} - 24 x + C

f (x) = \frac{5}{x}

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What is the integral of (x^2-6x+8)(x-3) ?
The integral of (x^2-6x+8)(x-3) is (x^4)/4-3x^3+13x^2-24x+C