What is the integral of (4x+9)(1-x) ?

The integral of (4x+9)(1-x) is -(4x^3)/3-(5x^2)/2+9x+C

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integral of (4x+9)(1-x)

\int (4 x + 9) (1 - x) d x

- \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 9 x + C

Solution steps

\int (4 x + 9) (1 - x) d x

Expand

(4 x + 9) (1 - x) : - 4 x^{2} - 5 x + 9

= \int - 4 x^{2} - 5 x + 9 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 5 x d x + \int 9 d x

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

\int 5 x d x = \frac{5 x ^{2}}{2}

\int 9 d x = 9 x

= - \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 9 x

Add a constant to the solution

= - \frac{4 x ^{3}}{3} - \frac{5 x ^{2}}{2} + 9 x + C

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What is the integral of (4x+9)(1-x) ?
The integral of (4x+9)(1-x) is -(4x^3)/3-(5x^2)/2+9x+C