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

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

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integral of (-3sqrt(x)+4x)

\int (- 3 x + 4 x) d x

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

Solution steps

\int - 3 x + 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 3 x d x + \int 4 x d x

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

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

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

Add a constant to the solution

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

d y - (y - 7)^{2} d x = 0

\int e^{4 x} cos (2 x) d x

\frac{\partial}{\partial y} ((sin (m x + n y))^{2})

\int 2 e^{- x^{2}} d x

\frac{d}{d x} (\frac{1}{6} (2 x + 1)^{3})

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