We are tasked with finding the integral of the function:

$\int \frac{3^{3\sqrt{x} - 2}}{2\sqrt{x}} \, dx$

Step-by-Step Solution:

Step 1: Substitution

Let’s first set a substitution to simplify the expression:

Let $u = \sqrt{x} = x^{1/2}$, so that: $\frac{du}{dx} = \frac{1}{2\sqrt{x}} \implies dx = 2u \, du$

This transforms $\sqrt{x}$ to $u$, and the integral becomes: $\int \frac{3^{3u - 2}}{2u} \cdot 2u \, du$

The $2u$ terms cancel out, and we are left with: $\int 3^{3u - 2} \, du$

Step 2: Break the exponent

We can express the exponent $3u - 2$ as: $3^{3u - 2} = \frac{3^{3u}}{3^2} = \frac{3^{3u}}{9}$

Thus, the integral becomes: $\frac{1}{9} \int 3^{3u} \, du$

Step 3: Another substitution

Now, let’s perform another substitution. Let $v = 3u$, so that: $dv = 3 \, du \implies du = \frac{dv}{3}$

Thus, the integral becomes: $\frac{1}{9} \int 3^v \cdot \frac{dv}{3} = \frac{1}{27} \int 3^v \, dv$

Step 4: Integrate

The integral of $3^v$ with respect to $v$ is: $\int 3^v \, dv = \frac{3^v}{\ln(3)}$

Thus, the result of the integral is: $\frac{1}{27} \cdot \frac{3^v}{\ln(3)} = \frac{3^v}{27\ln(3)}$

Step 5: Back-substitute $v$ and $u$

Now, substitute back $v = 3u$ and $u = \sqrt{x}$: $\frac{3^{3u}}{27\ln(3)} = \frac{3^{3\sqrt{x}}}{27\ln(3)}$

Thus, the final answer is: $\int \frac{3^{3\sqrt{x} - 2}}{2\sqrt{x}} \, dx = \frac{3^{3\sqrt{x}}}{27\ln(3)} + C$

where $C$ is the constant of integration.

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Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. What is the integral of $\frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx$?
2. Find the derivative of $3^{3\sqrt{x}}$.
3. How do you perform substitution for integrals involving square roots?
4. Solve $\int \frac{a^{\sqrt{x}}}{\sqrt{x}} \, dx$, where $a$ is a constant.
5. Evaluate $\int 3^{2x} \, dx$.

Tip: For integrals involving exponential functions, consider substitutions that simplify the exponent before integrating.