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

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

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integral of (3x-4)(x+3)

\int (3 x - 4) (x + 3) d x

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

Solution steps

\int (3 x - 4) (x + 3) d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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