What is the integral of 9(3x-2)^2 ?

The integral of 9(3x-2)^2 is 9(3x^3-6x^2+4x)+C

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integral of 9(3x-2)^2

\int 9 (3 x - 2)^{2} d x

9 (3 x^{3} - 6 x^{2} + 4 x) + C

Solution steps

\int 9 (3 x - 2)^{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int (3 x - 2)^{2} d x

Expand

(3 x - 2)^{2} : 9 x^{2} - 12 x + 4

= 9 \cdot \int 9 x^{2} - 12 x + 4 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

= 9 (\int 9 x^{2} d x - \int 12 x d x + \int 4 d x)

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

\int 12 x d x = 6 x^{2}

\int 4 d x = 4 x

= 9 (3 x^{3} - 6 x^{2} + 4 x)

Add a constant to the solution

= 9 (3 x^{3} - 6 x^{2} + 4 x) + C

\frac{d ^{3}}{d x ^{3}} (\frac{5}{x ^{2}})

f (x) = (x^{2} - 3 x + 8)^{3}

x \to - \infty lim (x^{2} + 1)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 5 x e^{4 x}

What is the integral of 9(3x-2)^2 ?
The integral of 9(3x-2)^2 is 9(3x^3-6x^2+4x)+C