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

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

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integral of (12x^3-6x^2+4x)

\int (12 x^{3} - 6 x^{2} + 4 x) d x

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

Solution steps

\int 12 x^{3} - 6 x^{2} + 4 x 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 12 x^{3} d x - \int 6 x^{2} d x + \int 4 x d x

\int 12 x^{3} d x = 3 x^{4}

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

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

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

Add a constant to the solution

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

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