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

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

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integral of (4x+3)(2x^2+3x)3

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

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

Solution steps

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

Take the constant out:

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

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

Expand

(4 x + 3) (2 x^{2} + 3 x) : 8 x^{3} + 18 x^{2} + 9 x

= 3 \cdot \int 8 x^{3} + 18 x^{2} + 9 x 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 8 x^{3} d x + \int 18 x^{2} d x + \int 9 x d x)

\int 8 x^{3} d x = 2 x^{4}

\int 18 x^{2} d x = 6 x^{3}

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

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

Add a constant to the solution

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

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