What is the integral of (x-2)(x^2-4x+9) ?

The integral of (x-2)(x^2-4x+9) is (x^4)/4-2x^3+(17x^2)/2-18x+C

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integral of (x-2)(x^2-4x+9)

\int (x - 2) (x^{2} - 4 x + 9) d x

\frac{x ^{4}}{4} - 2 x^{3} + \frac{17 x ^{2}}{2} - 18 x + C

Solution steps

\int (x - 2) (x^{2} - 4 x + 9) d x

Expand

(x - 2) (x^{2} - 4 x + 9) : x^{3} - 6 x^{2} + 17 x - 18

= \int x^{3} - 6 x^{2} + 17 x - 18 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 6 x^{2} d x + \int 17 x d x - \int 18 d x

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

\int 6 x^{2} d x = 2 x^{3}

\int 17 x d x = \frac{17 x ^{2}}{2}

\int 18 d x = 18 x

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{17 x ^{2}}{2} - 18 x

Add a constant to the solution

= \frac{x ^{4}}{4} - 2 x^{3} + \frac{17 x ^{2}}{2} - 18 x + C

\int 2 x^{2} e^{2 x} d x

\frac{\partial}{\partial s} (s t)

x^{4} - 2 x^{3} + 5

\int_{0}^{t} x d x

\frac{d}{d x} (cos (3 x^{7} + 2))

What is the integral of (x-2)(x^2-4x+9) ?
The integral of (x-2)(x^2-4x+9) is (x^4)/4-2x^3+(17x^2)/2-18x+C