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.