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

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

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integral of 4(x+3)^2

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

4 (\frac{x ^{3}}{3} + 3 x^{2} + 9 x) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int (x + 3)^{2} d x

Expand

(x + 3)^{2} : x^{2} + 6 x + 9

= 4 \cdot \int x^{2} + 6 x + 9 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

= 4 (\int x^{2} d x + \int 6 x d x + \int 9 d x)

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

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

\int 9 d x = 9 x

= 4 (\frac{x ^{3}}{3} + 3 x^{2} + 9 x)

Add a constant to the solution

= 4 (\frac{x ^{3}}{3} + 3 x^{2} + 9 x) + C

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