What is the integral of (\sqrt[3]{x})/2 ?

The integral of (\sqrt[3]{x})/2 is 3/8 x^{4/3}+C

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integral of (\sqrt[3]{x})/2

\frac{3}{8} x^{\frac{4}{3}} + C

Solution steps

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{3}{4} x^{\frac{4}{3}}

Simplify

\frac{1}{2} \cdot \frac{3}{4} x^{\frac{4}{3}} : \frac{3}{8} x^{\frac{4}{3}}

= \frac{3}{8} x^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{8} x^{\frac{4}{3}} + C

\frac{d ^{2}}{d x ^{2}} (7 x^{- 3} + 5 t^{- \frac{8}{3}})

\int_{0}^{1} (3 x)^{2} d x

\frac{4 s - 18}{s ^{2} - 6 s + 10}

y = x + 5

\frac{d y}{d x} + a y = 0

What is the integral of (\sqrt[3]{x})/2 ?
The integral of (\sqrt[3]{x})/2 is 3/8 x^{4/3}+C