What is the integral of \sqrt[3]{16sqrt(4x^3)} ?

The integral of \sqrt[3]{16sqrt(4x^3)} is (4\sqrt[3]{4})/3 x^{3/2}+C

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integral of \sqrt[3]{16sqrt(4x^3)}

\int 3 16 4 x^{3} d x

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

Solution steps

\int 3 16 4 x^{3} d x

Apply u-substitution

: \int \frac{2 3 4}{3 u} d u

= \int \frac{2 3 4}{3 u} d u

Take the constant out:

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

= \frac{2 3 4}{3} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = x^{3}

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of \sqrt[3]{16sqrt(4x^3)} ?
The integral of \sqrt[3]{16sqrt(4x^3)} is (4\sqrt[3]{4})/3 x^{3/2}+C