What is the integral of 3(2x+1)^2 ?

The integral of 3(2x+1)^2 is 4x^3+6x^2+3x+C

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integral of 3(2x+1)^2

\int 3 (2 x + 1)^{2} d x

4 x^{3} + 6 x^{2} + 3 x + C

Solution steps

\int 3 (2 x + 1)^{2} d x

Take the constant out:

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

= 3 \cdot \int (2 x + 1)^{2} d x

Expand

(2 x + 1)^{2} : 4 x^{2} + 4 x + 1

= 3 \cdot \int 4 x^{2} + 4 x + 1 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 4 x^{2} d x + \int 4 x d x + \int 1 d x)

\int 4 x^{2} d x = \frac{4 x ^{3}}{3}

\int 4 x d x = 2 x^{2}

\int 1 d x = x

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

Simplify

3 (\frac{4 x ^{3}}{3} + 2 x^{2} + x) : 4 x^{3} + 6 x^{2} + 3 x

= 4 x^{3} + 6 x^{2} + 3 x

Add a constant to the solution

= 4 x^{3} + 6 x^{2} + 3 x + C

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