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

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

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

\int \frac{x}{3 1 - x ^{2}} d x

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

Solution steps

\int \frac{x}{3 1 - 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 = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{3 x}{1 - 4 x ^{2}} d x

\int \frac{cos ( x )}{2 - sin ^{2} ( x )} d x

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

\int 2 t + 7 d t

\int \frac{4 x}{4 x ^{2} + 1} d x

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