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

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

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integral of x/(\sqrt[3]{1-3x^2)}

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

Solution steps

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{1}{6} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = 1 - 3 x^{2}

= - \frac{1}{6} \cdot \frac{3}{2} (1 - 3 x^{2})^{\frac{2}{3}}

Simplify

- \frac{1}{6} \cdot \frac{3}{2} (1 - 3 x^{2})^{\frac{2}{3}} : - \frac{1}{4} (1 - 3 x^{2})^{\frac{2}{3}}

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

Add a constant to the solution

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

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