What is the integral of (10\sqrt[3]{x^2}) ?

The integral of (10\sqrt[3]{x^2}) is 6x^{5/3}+C

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integral of (10\sqrt[3]{x^2})

6 x^{\frac{5}{3}} + C

Solution steps

Take the constant out:

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

assuming

x \geq 0

= 10 \cdot \int x^{\frac{2}{3}} d x

Apply the Power Rule

: \frac{3}{5} x^{\frac{5}{3}}

= 10 \cdot \frac{3}{5} x^{\frac{5}{3}}

Simplify

10 \cdot \frac{3}{5} x^{\frac{5}{3}} : 6 x^{\frac{5}{3}}

= 6 x^{\frac{5}{3}}

Add a constant to the solution

= 6 x^{\frac{5}{3}} + C

\frac{\partial}{\partial x} (x e^{4 x})

2 x = y, x + y = 4, x + y = 12

\int \frac{( 3 x ^{2} + 2 x - 3 )}{x ^{3} - x} d x

t^{2} e^{- t} + (ln (t))^{2}

y = x^{2}, y = \frac{2}{x ^{2} + 1}

What is the integral of (10\sqrt[3]{x^2}) ?
The integral of (10\sqrt[3]{x^2}) is 6x^{5/3}+C