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

The integral of (x^3)/(\sqrt[3]{3-2x^2)} is 1/8 (3/5 (3-2x^2)^{5/3}-9/2 (3-2x^2)^{2/3})+C

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

\frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{2}{3}} d u = \frac{3}{5} u^{\frac{5}{3}}

= \frac{1}{8} (\frac{3}{5} u^{\frac{5}{3}} - \frac{9}{2} u^{\frac{2}{3}})

Substitute back

u = 3 - 2 x^{2}

= \frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 x^{2})^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{8} (\frac{3}{5} (3 - 2 x^{2})^{\frac{5}{3}} - \frac{9}{2} (3 - 2 x^{2})^{\frac{2}{3}}) + C

x \to - 2 lim (\frac{3 x - 9}{x + 1})

n = 2 \sum \infty \frac{1}{( - 5 ) ^{n}}

n \to + 0 - lim (5 + \frac{4}{n})

\int 3 \cdot cos (3 x) d x

\int \frac{x ^{15}}{x ^{16} + 1} d x

What is the integral of (x^3)/(\sqrt[3]{3-2x^2)} ?
The integral of (x^3)/(\sqrt[3]{3-2x^2)} is 1/8 (3/5 (3-2x^2)^{5/3}-9/2 (3-2x^2)^{2/3})+C