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

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

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

\frac{3}{4} x^{\frac{4}{3}} - 2 x + C

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 2 d x = 2 x

= \frac{3}{4} x^{\frac{4}{3}} - 2 x

Add a constant to the solution

= \frac{3}{4} x^{\frac{4}{3}} - 2 x + C

y = \frac{( 3 x ^{2} + 4 ) ( 2 x + 1 )}{5 x - 9}

f (z) = \frac{z - 8}{z + 8}

x \to 3 lim (\frac{5}{x ^{2} - 9})

x \to 0 + lim (x^{4 x})

\frac{d y}{d x} = \frac{1}{5} x

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