The integral of the function \(\frac{e^{2x}}{e^{4x} + 9}\) can be solved using substitution.

### Step 1: Simplify the expression
The given expression can be simplified as:
\[
\frac{e^{2x}}{e^{4x} + 9} = \frac{e^{2x}}{(e^{2x})^2 + 9}
\]
Let \( u = e^{2x} \), so \( du = 2e^{2x} dx \), and \( dx = \frac{du}{2e^{2x}} \).

### Step 2: Substitute into the integral
This changes the integral into:
\[
\int \frac{e^{2x}}{e^{4x} + 9} dx = \int \frac{1}{u^2 + 9} \cdot \frac{du}{2}
\]

### Step 3: Solve the integral
The integral now becomes:
\[
\frac{1}{2} \int \frac{du}{u^2 + 9}
\]
This is a standard integral, with the formula:
\[
\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan \left( \frac{u}{a} \right)
\]
where \(a = 3\). Applying this formula, we get:
\[
\frac{1}{2} \cdot \frac{1}{3} \arctan \left( \frac{u}{3} \right) + C = \frac{1}{6} \arctan \left( \frac{e^{2x}}{3} \right) + C
\]

### Final Answer:
\[
\int \frac{e^{2x}}{e^{4x} + 9} dx = \frac{1}{6} \arctan \left( \frac{e^{2x}}{3} \right) + C
\]

Let me know if you would like more details on any part of the solution or have further questions.

Here are 5 related questions for further exploration:
1. How would you approach \(\int \frac{e^{kx}}{e^{2kx} + c} dx\) for different constants \(k\) and \(c\)?
2. Can you solve \(\int \frac{e^x}{e^{2x} + 1} dx\)?
3. How would you solve \(\int \frac{dx}{x^2 + a^2}\)?
4. What happens if the integral has limits, such as \(\int_0^1 \frac{e^{2x}}{e^{4x} + 9} dx\)?
5. Can you derive the arctangent formula used for integration from first principles?

**Tip:** Whenever you're working with exponential terms, substitution is often a useful method to simplify the integrand.

Discover how to solve the integral of e^(2x) / (e^(4x) + 9) using the substitution method. This problem explores advanced integration techniques and the arctangent formula, ideal for students in Grades 11-12.

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