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

The integral of (3-4x)(-x-5) is (4x^3)/3+(17x^2)/2-15x+C

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

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

\frac{4 x ^{3}}{3} + \frac{17 x ^{2}}{2} - 15 x + C

Solution steps

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

Expand

(3 - 4 x) (- x - 5) : 4 x^{2} + 17 x - 15

= \int 4 x^{2} + 17 x - 15 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 4 x^{2} d x + \int 17 x d x - \int 15 d x

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

\int 17 x d x = \frac{17 x ^{2}}{2}

\int 15 d x = 15 x

= \frac{4 x ^{3}}{3} + \frac{17 x ^{2}}{2} - 15 x

Add a constant to the solution

= \frac{4 x ^{3}}{3} + \frac{17 x ^{2}}{2} - 15 x + C

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What is the integral of (3-4x)(-x-5) ?
The integral of (3-4x)(-x-5) is (4x^3)/3+(17x^2)/2-15x+C