What is the integral of (x+4)(3x-5) ?

The integral of (x+4)(3x-5) is x^3+(7x^2)/2-20x+C

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integral of (x+4)(3x-5)

\int (x + 4) (3 x - 5) d x

x^{3} + \frac{7 x ^{2}}{2} - 20 x + C

Solution steps

\int (x + 4) (3 x - 5) d x

Expand

(x + 4) (3 x - 5) : 3 x^{2} + 7 x - 20

= \int 3 x^{2} + 7 x - 20 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 3 x^{2} d x + \int 7 x d x - \int 20 d x

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

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 20 d x = 20 x

= x^{3} + \frac{7 x ^{2}}{2} - 20 x

Add a constant to the solution

= x^{3} + \frac{7 x ^{2}}{2} - 20 x + C

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What is the integral of (x+4)(3x-5) ?
The integral of (x+4)(3x-5) is x^3+(7x^2)/2-20x+C