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

x^{- \frac{3}{4}}

\int_{0}^{\frac{π}{6}} (9 sec^{2} (x)) d x

\int \frac{1}{7 - 3 x} d x

y^{^{'}} = 9 x + y

(2 e^{y} - x) (\frac{d y}{d x}) = 1

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