What is the integral of x^2+2x-\sqrt[3]{3x} ?

The integral of x^2+2x-\sqrt[3]{3x} is (x^3)/3+x^2-\sqrt[3]{3}x\sqrt[3]{x}+3/(4\sqrt[3]{9)}x^{4/3}+C

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integral of x^2+2x-\sqrt[3]{3x}

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

Simplify

Add a constant to the solution

f (x) = 5 e^{x} cos (x)

\int_{- 6}^{x} 36 - t^{2} d t

(1 + x)^{\frac{1}{2}}, 0

y^{^{'}} = (y + 5) (x + 2), y (0) = - 1

h \to 0 lim (\frac{1 - ( z + h ) \cdot 2 ( z + h ) - 1 - z \cdot 2 z ^{- 1}}{h})

What is the integral of x^2+2x-\sqrt[3]{3x} ?
The integral of x^2+2x-\sqrt[3]{3x} is (x^3)/3+x^2-\sqrt[3]{3}x\sqrt[3]{x}+3/(4\sqrt[3]{9)}x^{4/3}+C