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

The integral of 3(3x+4)^2 is 3(3x^3+12x^2+16x)+C

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integral of 3(3x+4)^2

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

3 (3 x^{3} + 12 x^{2} + 16 x) + C

Solution steps

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

Take the constant out:

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

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

Expand

(3 x + 4)^{2} : 9 x^{2} + 24 x + 16

= 3 \cdot \int 9 x^{2} + 24 x + 16 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

= 3 (\int 9 x^{2} d x + \int 24 x d x + \int 16 d x)

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

\int 24 x d x = 12 x^{2}

\int 16 d x = 16 x

= 3 (3 x^{3} + 12 x^{2} + 16 x)

Add a constant to the solution

= 3 (3 x^{3} + 12 x^{2} + 16 x) + C

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What is the integral of 3(3x+4)^2 ?
The integral of 3(3x+4)^2 is 3(3x^3+12x^2+16x)+C