We are tasked with evaluating the integral:

$\int \frac{-6\sqrt{x}}{1 + x^3} \, dx$

Step 1: Substitution to simplify the integrand

Let’s first introduce a substitution that might simplify the problem. Since we have a term involving $x^3$ in the denominator and a square root of $x$ in the numerator, we can use a substitution that could reduce these complexities.

Let: $u = x^{3/2}$ This gives us: $du = \frac{3}{2} x^{1/2} dx \quad \text{or equivalently} \quad dx = \frac{2}{3} \frac{du}{x^{1/2}}$

Now, since $x^{1/2} = \sqrt{x}$, we rewrite the integral as:

$\int \frac{-6 \sqrt{x}}{1 + x^3} \, dx$

Substituting $dx = \frac{2}{3} \frac{du}{x^{1/2}}$ and $x^{3/2} = u$,