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

The integral of 4sqrt(x)-6x^2+3 is 8/3 x^{3/2}-2x^3+3x+C

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

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

\frac{8}{3} x^{\frac{3}{2}} - 2 x^{3} + 3 x + C

Solution steps

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

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

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

\int 3 d x = 3 x

= \frac{8}{3} x^{\frac{3}{2}} - 2 x^{3} + 3 x

Add a constant to the solution

= \frac{8}{3} x^{\frac{3}{2}} - 2 x^{3} + 3 x + C

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What is the integral of 4sqrt(x)-6x^2+3 ?
The integral of 4sqrt(x)-6x^2+3 is 8/3 x^{3/2}-2x^3+3x+C