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

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

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integral of x/(\sqrt[3]{4-x^2)}

\int \frac{x}{3 4 - x ^{2}} d x

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

Solution steps

\int \frac{x}{3 4 - x ^{2}} d x

Apply u-substitution

: \int - \frac{1}{2 3 u} d u

= \int - \frac{1}{2 3 u} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

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

Substitute back

u = 4 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int coth (\frac{x}{4}) d x

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