What is the integral of (2x)/(\sqrt[4]{1-x^2)} ?

The integral of (2x)/(\sqrt[4]{1-x^2)} is -4/3 (1-x^2)^{3/4}+C

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integral of (2x)/(\sqrt[4]{1-x^2)}

- \frac{4}{3} (1 - x^{2})^{\frac{3}{4}} + C

Solution steps

Take the constant out:

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

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

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

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x y - 5)^{2})

x \to 0 lim (\frac{( e ) ^{x}}{x})

\frac{d}{d x} ((3 x - 4) (x^{3} + 5))

\int - sec^{3} (x) d x

f (x)^{2}

What is the integral of (2x)/(\sqrt[4]{1-x^2)} ?
The integral of (2x)/(\sqrt[4]{1-x^2)} is -4/3 (1-x^2)^{3/4}+C