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

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

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

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

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

Solution steps

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

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

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

\int 4 d x = 4 x

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

Add a constant to the solution

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

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