What is the integral of-6x^3+9x^2+4x-3 ?

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

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integral of-6x^3+9x^2+4x-3

\int - 6 x^{3} + 9 x^{2} + 4 x - 3 d x

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

Solution steps

\int - 6 x^{3} + 9 x^{2} + 4 x - 3 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 6 x^{3} d x + \int 9 x^{2} d x + \int 4 x d x - \int 3 d x

\int 6 x^{3} d x = \frac{3 x ^{4}}{2}

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

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

\int 3 d x = 3 x

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

Add a constant to the solution

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

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