What is the integral of (8x^3-6x^2+12x-9) ?

The integral of (8x^3-6x^2+12x-9) is 2x^4-2x^3+6x^2-9x+C

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integral of (8x^3-6x^2+12x-9)

\int (8 x^{3} - 6 x^{2} + 12 x - 9) d x

2 x^{4} - 2 x^{3} + 6 x^{2} - 9 x + C

Solution steps

\int 8 x^{3} - 6 x^{2} + 12 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 8 x^{3} d x - \int 6 x^{2} d x + \int 12 x d x - \int 9 d x

\int 8 x^{3} d x = 2 x^{4}

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

\int 12 x d x = 6 x^{2}

\int 9 d x = 9 x

= 2 x^{4} - 2 x^{3} + 6 x^{2} - 9 x

Add a constant to the solution

= 2 x^{4} - 2 x^{3} + 6 x^{2} - 9 x + C

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What is the integral of (8x^3-6x^2+12x-9) ?
The integral of (8x^3-6x^2+12x-9) is 2x^4-2x^3+6x^2-9x+C