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

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

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integral of 3sqrt(x)-6x

\int 3 x - 6 x d x

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

Solution steps

\int 3 x - 6 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 3 x d x - \int 6 x d x

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

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 3)^{7})

\int_{0}^{5.5} 120 x d x

\int (9 x^{2}) d x

x^{2} sin (2 x)

x \to 7 lim (sin (\frac{π x}{2}))

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