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

The integral of (2x-3)(4x^2+6x+9) is 2x^4-27x+C

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integral of (2x-3)(4x^2+6x+9)

\int (2 x - 3) (4 x^{2} + 6 x + 9) d x

2 x^{4} - 27 x + C

Solution steps

\int (2 x - 3) (4 x^{2} + 6 x + 9) d x

Expand

(2 x - 3) (4 x^{2} + 6 x + 9) : 8 x^{3} - 27

= \int 8 x^{3} - 27 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

= \int 8 x^{3} d x - \int 27 d x

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

\int 27 d x = 27 x

= 2 x^{4} - 27 x

Add a constant to the solution

= 2 x^{4} - 27 x + C

\int \frac{x ^{3}}{x ^{4} - 4} d x

\int \frac{1}{( 5 - 2 x ) ^{2}} d x

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\int \frac{1}{tan ^{4} ( x )} d x

\int \frac{1}{2 + 2 x} d x

What is the integral of (2x-3)(4x^2+6x+9) ?
The integral of (2x-3)(4x^2+6x+9) is 2x^4-27x+C