What is the integral of (6x+12)(x-3) ?

The integral of (6x+12)(x-3) is 2x^3-3x^2-36x+C

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integral of (6x+12)(x-3)

\int (6 x + 12) (x - 3) d x

2 x^{3} - 3 x^{2} - 36 x + C

Solution steps

\int (6 x + 12) (x - 3) d x

Apply Integration By Parts

: (6 x + 12) (\frac{x ^{2}}{2} - 3 x) - \int 6 (\frac{x ^{2}}{2} - 3 x) d x

= (6 x + 12) (\frac{x ^{2}}{2} - 3 x) - \int 6 (\frac{x ^{2}}{2} - 3 x) d x

\int 6 (\frac{x ^{2}}{2} - 3 x) d x = x^{3} - 9 x^{2}

= (6 x + 12) (\frac{x ^{2}}{2} - 3 x) - (x^{3} - 9 x^{2})

Simplify

(6 x + 12) (\frac{x ^{2}}{2} - 3 x) - (x^{3} - 9 x^{2}) : 2 x^{3} - 3 x^{2} - 36 x

= 2 x^{3} - 3 x^{2} - 36 x

Add a constant to the solution

= 2 x^{3} - 3 x^{2} - 36 x + C

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What is the integral of (6x+12)(x-3) ?
The integral of (6x+12)(x-3) is 2x^3-3x^2-36x+C