What is the integral of 6(3x+1)^2 ?

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

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integral of 6(3x+1)^2

\int 6 (3 x + 1)^{2} d x

6 (3 x^{3} + 3 x^{2} + x) + C

Solution steps

\int 6 (3 x + 1)^{2} d x

Take the constant out:

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

= 6 \cdot \int (3 x + 1)^{2} d x

Expand

(3 x + 1)^{2} : 9 x^{2} + 6 x + 1

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

= 6 (\int 9 x^{2} d x + \int 6 x d x + \int 1 d x)

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

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

\int 1 d x = x

= 6 (3 x^{3} + 3 x^{2} + x)

Add a constant to the solution

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

What is the integral of 6(3x+1)^2 ?
The integral of 6(3x+1)^2 is 6(3x^3+3x^2+x)+C