What is the integral of (9x+2)(3x-5) ?

The integral of (9x+2)(3x-5) is 9x^3-(39x^2)/2-10x+C

integral of (9x+2)(3x-5)

\int (9 x + 2) (3 x - 5) d x

9 x^{3} - \frac{39 x ^{2}}{2} - 10 x + C

Solution steps

\int (9 x + 2) (3 x - 5) d x

Expand

(9 x + 2) (3 x - 5) : 27 x^{2} - 39 x - 10

= \int 27 x^{2} - 39 x - 10 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 27 x^{2} d x - \int 39 x d x - \int 10 d x

\int 27 x^{2} d x = 9 x^{3}

\int 39 x d x = \frac{39 x ^{2}}{2}

\int 10 d x = 10 x

= 9 x^{3} - \frac{39 x ^{2}}{2} - 10 x

Add a constant to the solution

= 9 x^{3} - \frac{39 x ^{2}}{2} - 10 x + C

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What is the integral of (9x+2)(3x-5) ?
The integral of (9x+2)(3x-5) is 9x^3-(39x^2)/2-10x+C