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

The integral of (x^3)/(\sqrt[3]{x^2+4)} is 1/10 (3(x^2+4)^{5/3}-30(x^2+4)^{2/3})+C

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

\frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\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 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 v^{\frac{2}{3}} d v = \frac{3}{5} v^{\frac{5}{3}}

= \frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} v^{\frac{5}{3}} - 6 v^{\frac{2}{3}})

Substitute back

= \frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{5}{3}} - 6 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{2}{3}})

Simplify

\frac{1}{3} \cdot \frac{3}{2} (\frac{3}{5} ((x^{3})^{\frac{2}{3}} + 4)^{\frac{5}{3}} - 6 ((x^{3})^{\frac{2}{3}} + 4)^{\frac{2}{3}}) : \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}})

= \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}})

Add a constant to the solution

= \frac{1}{10} (3 (x^{2} + 4)^{\frac{5}{3}} - 30 (x^{2} + 4)^{\frac{2}{3}}) + C

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What is the integral of (x^3)/(\sqrt[3]{x^2+4)} ?
The integral of (x^3)/(\sqrt[3]{x^2+4)} is 1/10 (3(x^2+4)^{5/3}-30(x^2+4)^{2/3})+C