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The given integral is:

\[
\int \frac{3^{2x}}{1 + 3^{2x}} \, dx
\]

To solve this, let's approach it step by step:

### Step 1: Substitution
We will make a substitution to simplify the expression. Let:

\[
u = 3^{2x}
\]

Now, differentiate \(u\) with respect to \(x\):

\[
\frac{du}{dx} = 2 \cdot 3^{2x} \ln(3)
\]

Thus, solving for \(dx\):

\[
dx = \frac{du}{2 \cdot u \ln(3)}
\]

### Step 2: Substitute into the integral
Now substitute \(u = 3^{2x}\) and \(dx = \frac{du}{2 \cdot u \ln(3)}\) into the original integral:

\[
\int \frac{u}{1 + u} \cdot \frac{du}{2 \cdot u \ln(3)}
\]

This simplifies to:

\[
\frac{1}{2 \ln(3)} \int \frac{du}{1 + u}
\]

### Step 3: Solve the integral
The integral of \( \frac{1}{1+u} \) is a standard result:

\[
\int \frac{1}{1+u} \, du = \ln|1+u|
\]

Thus, the solution to the integral is:

\[
\frac{1}{2 \ln(3)} \ln|1 + u| + C
\]

### Step 4: Substitute back \(u = 3^{2x}\)
Finally, substitute \(u = 3^{2x}\) back into the expression:

\[
\frac{1}{2 \ln(3)} \ln|1 + 3^{2x}| + C
\]

Thus, the final solution is:

\[
\frac{1}{2 \ln(3)} \ln(1 + 3^{2x}) + C
\]

### Do you want any further clarifications or detailed steps?

Here are five relative questions:
1. What other types of substitution methods can be used for integrals?
2. How does the logarithmic function help in solving integrals like this?
3. Why is the natural logarithm involved when differentiating or integrating exponential functions?
4. What are the conditions for applying substitution in integrals?
5. Can you apply partial fractions for integrals involving more complex rational functions?

**Tip:** When dealing with exponentials in integrals, substitution can often simplify the problem by converting the expression into a more manageable form.

Learn how to solve the integral ∫ (3^(2x)) / (1 + 3^(2x)) dx using the substitution method. This step-by-step solution covers key concepts in calculus, including the substitution rule and logarithmic integration.

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