What is the integral of (-x^3+6x^2-9x) ?

The integral of (-x^3+6x^2-9x) is -(x^4)/4+2x^3-(9x^2)/2+C

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integral of (-x^3+6x^2-9x)

\int (- x^{3} + 6 x^{2} - 9 x) d x

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

Solution steps

\int - x^{3} + 6 x^{2} - 9 x 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 x^{3} d x + \int 6 x^{2} d x - \int 9 x d x

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

\int 6 x^{2} d x = 2 x^{3}

\int 9 x d x = \frac{9 x ^{2}}{2}

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

Add a constant to the solution

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

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What is the integral of (-x^3+6x^2-9x) ?
The integral of (-x^3+6x^2-9x) is -(x^4)/4+2x^3-(9x^2)/2+C