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

The integral of (x^5)/(\sqrt[3]{2x^3+1)} is 1/40 (4x^3-3)(2x^3+1)^{2/3}+C

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integral of (x^5)/(\sqrt[3]{2x^3+1)}

\frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\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}{12} (\frac{3}{5} u^{\frac{5}{3}} - \frac{3}{2} u^{\frac{2}{3}})

Substitute back

u = 2 x^{3} + 1

= \frac{1}{12} (\frac{3}{5} (2 x^{3} + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x^{3} + 1)^{\frac{2}{3}})

Simplify

\frac{1}{12} (\frac{3}{5} (2 x^{3} + 1)^{\frac{5}{3}} - \frac{3}{2} (2 x^{3} + 1)^{\frac{2}{3}}) : \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}}

= \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}}

Add a constant to the solution

= \frac{1}{40} (4 x^{3} - 3) (2 x^{3} + 1)^{\frac{2}{3}} + C

What is the integral of (x^5)/(\sqrt[3]{2x^3+1)} ?
The integral of (x^5)/(\sqrt[3]{2x^3+1)} is 1/40 (4x^3-3)(2x^3+1)^{2/3}+C