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

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

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integral of (6x^2-3x)

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

2 x^{3} - \frac{3 x ^{2}}{2} + C

Solution steps

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

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

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

= 2 x^{3} - \frac{3 x ^{2}}{2}

Add a constant to the solution

= 2 x^{3} - \frac{3 x ^{2}}{2} + C

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