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

x \to 0 lim (\frac{2 x ^{2} - x}{x})

y = (x)^{x}

f (x) = 7 sin (x)

\frac{d}{d x} (\frac{- 2}{x})

\frac{d y}{d x} = 3 + 2 y

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