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

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

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integral of (8x^3-3x^2+8)

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

2 x^{4} - x^{3} + 8 x + C

Solution steps

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

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

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

\int 8 d x = 8 x

= 2 x^{4} - x^{3} + 8 x

Add a constant to the solution

= 2 x^{4} - x^{3} + 8 x + C

\frac{d}{d x} (x^{5} e^{6 x})

x \to \infty lim (\frac{2}{x})

\frac{d}{d x} (- 2 cos (3 x))

x \to 0 + lim (\frac{x}{3 x} - 7 x)

\int \frac{3}{91} x^{2} d x

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