What is the integral of 8x^3-9x^2+4 ?

The integral of 8x^3-9x^2+4 is 2x^4-3x^3+4x+C

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integral of 8x^3-9x^2+4

\int 8 x^{3} - 9 x^{2} + 4 d x

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

Solution steps

\int 8 x^{3} - 9 x^{2} + 4 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 9 x^{2} d x + \int 4 d x

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

\int (8 x + 3)^{4} d x

\frac{d ^{2}}{d x ^{2}} (e^{x} - x^{4})

(48 - 2 x) - x, [- 8, 24]

x \to \infty lim (\frac{- 5}{x})

\frac{1}{( s - 3 )}

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