The value of $\int \frac{x^5}{\sqrt{1 + x^3}} dx$ is

Looking at the integral $\int \frac{x^5}{\sqrt{1 + x^3}} dx$, the expression $x^3$ inside the square root suggests a substitution.

Let $u = 1 + x^3$

Then $du = 3x^2 dx$, which gives:

$x^2 dx = \frac{du}{3}$

From the substitution: $x^3 = u - 1$

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Rewrite $x^5$ as:

$x^5 = x^3 \cdot x^2 = (u-1) \cdot x^2$

Substituting into the integral:

$\int \frac{x^5}{\sqrt{1 + x^3}} dx = \int \frac{(u-1) \cdot x^2}{\sqrt{u}} dx$

$= \int \frac{u-1}{\sqrt{u}} \cdot \frac{du}{3}$

$= \frac{1}{3} \int \frac{u-1}{\sqrt{u}} du$

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Simplify the integrand:

$\frac{1}{3} \int \frac{u-1}{\sqrt{u}} du = \frac{1}{3} \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) du$

$= \frac{1}{3} \int \left(u^{1/2} - u^{-1/2}\right) du$

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Apply the power rule:

$= \frac{1}{3} \left[\frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2}\right] + c$

$= \frac{1}{3} \left[\frac{2u^{3/2}}{3} - 2u^{1/2}\right] + c$

$= \frac{2u^{3/2}}{9} - \frac{2u^{1/2}}{3} + c$

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Substitute back $u = 1 + x^3$:

$\int \frac{x^5}{\sqrt{1 + x^3}} dx = \frac{2}{9}(1 + x^3)^{3/2} - \frac{2}{3}(1 + x^3)^{1/2} + c$