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

The integral of x/(\sqrt[3]{2-4x^2)} is -3/(8\sqrt[3]{2)}(1-2x^2)^{2/3}+C

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

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

Substitute back

u = 1 - 2 x^{2}

Simplify

Add a constant to the solution

\int arctan (a x) d x

x \to 3 lim (5^{2})

y = x^{2} + 2, y = x + 14

\int (x^{2} + x) d x

x \to - 2 + lim (\frac{x - 1}{x ^{2} - 4})

What is the integral of x/(\sqrt[3]{2-4x^2)} ?
The integral of x/(\sqrt[3]{2-4x^2)} is -3/(8\sqrt[3]{2)}(1-2x^2)^{2/3}+C