To solve the given integral using the substitution $u = x^3 + 19$, let's follow the substitution steps.

Step-by-Step Solution:

1. Substitute $u = x^3 + 19$: Differentiating $u$ with respect to $x$: $\frac{du}{dx} = 3x^2 \quad \Rightarrow \quad du = 3x^2 dx \quad \Rightarrow \quad x^2 dx = \frac{du}{3}.$

2. Rewrite the integral: Substitute $u = x^3 + 19$ and $x^2 dx = \frac{du}{3}$: $\int x^2 \sqrt{x^3 + 19} \, dx = \int x^2 \sqrt{u} \, \frac{du}{3}.$

   Simplify: $\int x^2 \sqrt{x^3 + 19} \, dx = \frac{1}{3} \int \sqrt{u} \, du.$

3. Solve the integral: The integral of $\sqrt{u}$ is: $\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2}.$

   Multiply by $\frac{1}{3}$: $\frac{1}{3} \int \sqrt{u} \, du = \frac{1}{3} \cdot \frac{2}{3} u^{3/2} = \frac{2}{9} u^{3/2}.$

4. Back-substitute $u = x^3 + 19$: Replace $u$ with $x^3 + 19$: $\frac{2}{9} u^{3/2} = \frac{2}{9} (x^3 + 19)^{3/2}.$

5. Add the constant of integration: $\int x^2 \sqrt{x^3 + 19} \, dx = \frac{2}{9} (x^3 + 19)^{3/2} + C.$

Final Answer:

$\int x^2 \sqrt{x^3 + 19} \, dx = \frac{2}{9} (x^3 + 19)^{3/2} + C.$

Would you like more details or clarification?

Here are 5 questions that build on this:

1. Why is substitution $u = x^3 + 19$ a useful technique in this problem?
2. How do we calculate the derivative $du$ and identify the replacement for $x^2 dx$?
3. What is the rule for integrating powers of $u$, such as $u^{1/2}$?
4. Why do we back-substitute $u = x^3 + 19$ in the final answer?
5. Can substitution methods be applied to integrals with other radicals or powers?

Tip: When using substitution, always express all terms of the integral in terms of the new variable $u$ before integrating.