91Ó°ÊÓ

We will identify the substitution to simplify the integral. We note that the denominator has the square root of the form \((1 + x^3)\). A good potential substitution is to let \(u = 1 + x^3\) to simplify the expression.

Now, we perform the substitution. Let \(u = 1 + x^3\). Differentiating both sides with respect to \(x\), we get $$ \frac{du}{dx} = 3x^2. $$ Now, we can express \(dx = \frac{du}{3x^2}\). We can also rewrite the integral in terms of \(u\): $$x^4 = u - 1$$, and our integral becomes, $$ \int \frac{x^{5} dx}{(u)^{1/2}} = \int \frac{(u - 1)x}{(u)^{1/2}} \cdot \frac{du}{3x^2}. $$

We can simplify the integral as $$ \int \frac{(u - 1) x}{(u)^{1/2}} \cdot \frac{du}{3x^2} = \frac{1}{3} \int \frac{u - 1}{u^{1/2}} du. $$ Now, split the integral into two parts: $$ \frac{1}{3} \int \frac{u - 1}{u^{1/2}} du = \frac{1}{3} \left( \int \frac{u}{u^{1/2}} du - \int \frac{1}{u^{1/2}} du \right). $$ Now, we can evaluate integrals for each part: $$ \int \frac{u}{u^{1/2}} du = \int u^{1/2} du, $$ $$ \int \frac{1}{u^{1/2}} du = \int u^{-1/2} du. $$ Integrating these, we get: $$ \int u^{1/2} du = \frac{2}{3} u^{3/2} + C_1, $$ $$ \int u^{-1/2} du = 2 u^{1/2} + C_2. $$ Combining these results, we get $$ \frac{1}{3} \left( \int \frac{u - 1}{u^{1/2}} du \right) = \frac{1}{3} \left( \frac{2}{3} u^{3/2} + C_1 - (2u^{1/2} + C_2) \right). $$ Simplifying the expression, we have $$ \frac{1}{3} \left( \int \frac{u - 1}{u^{1/2}} du \right) = \frac{2}{9} u^{3/2} - \frac{2}{3}u^{1/2} + C, $$ where \(C = \frac{1}{3}(C_1 - C_2)\).

Finally, we substitute back the original variable \(x\) by replacing \(u\) with \(1 + x^3\). Our final answer is $$ \int \frac{x^{5} dx}{(1+x^{3})^{1/2}} = \frac{2}{9}(1+x^{3})^{3/2} -\frac{2}{3}(1+x^{3})^{1/2} + C. $$