The integral provided is:

$\int 3e^{3\sqrt{x}} \, dx$

To integrate this expression, we'll first notice that it's more straightforward to perform a substitution rather than using integration by parts.

Step 1: Substitution

Let's substitute: $u = 3\sqrt{x}$ Then, $\frac{du}{dx} = \frac{3}{2\sqrt{x}} \implies du = \frac{3}{2\sqrt{x}} \, dx$ Thus, $dx = \frac{2\sqrt{x}}{3} \, du$

Since $u = 3\sqrt{x}$, we have $\sqrt{x} = \frac{u}{3}$. Substituting in $dx$:

$dx = \frac{2\sqrt{x}}{3} \, du = \frac{2 \cdot \frac{u}{3}}{3} \, du = \frac{2u}{9} \, du$

Substitute this back into the integral:

$\int 3e^{3\sqrt{x}} \, dx = \int 3e^u \cdot \frac{2u}{9} \, du = \frac{6}{9} \int u e^u \, du = \frac{2}{3} \int u e^u \, du$

Step 2: Integration by Parts

Now, apply integration by parts on $\int u e^u \, du$. Let: $u_1 = u, \quad dv = e^u \, du$ Then, $du_1 = du, \quad v = e^u$

Apply the integration by parts formula: $\int u e^u \, du = u e^u - \int e^u \, du$

Simplify the integral: $\int u e^u \, du = u e^u - e^u + C_1 = e^u(u - 1) + C_1$

Step 3: Substitute Back

Now, substituting back $u = 3\sqrt{x}$:

$\frac{2}{3} \int u e^u \, du = \frac{2}{3} \left[ e^{3\sqrt{x}}(3\sqrt{x} - 1) \right] + C_2$

So the final answer is:

$\int 3e^{3\sqrt{x}} \, dx = 2e^{3\sqrt{x}} \left(\sqrt{x} - \frac{1}{3}\right) + C$

where $C$ is the constant of integration.

Would you like more details on any part of this solution, or do you have any questions?

Here are some related questions:

1. How would you solve $\int e^{\sqrt{x}} \, dx$?
2. What substitution would you use for the integral $\int 2x e^{2x} \, dx$?
3. Can you integrate $\int e^{ax + b} \, dx$ directly?
4. What is the process for reversing a substitution in an integral?
5. How does integration by parts compare to substitution for different types of integrals?

Tip: Always look for substitution opportunities first before applying integration by parts, especially when dealing with functions that involve square roots or other compositions.