To solve the given integral:

$L = 2 \int_{0}^{\ln 3} \sqrt{e^{2x} + 1 + \frac{e^{-2x}}{4}} e^x \, dx$

we first simplify the integrand. Let's start by simplifying the expression inside the square root:

$e^{2x} + 1 + \frac{e^{-2x}}{4}$

We can rewrite this expression to make it easier to handle:

$e^{2x} + 1 + \frac{e^{-2x}}{4} = e^{2x} + 1 + \frac{1}{4e^{2x}}$

To simplify further, notice that if we multiply the integrand by $e^x$, it becomes:

$e^x \sqrt{e^{2x} + 1 + \frac{e^{-2x}}{4}}$

We can re-express the term inside the square root in a more convenient form:

$e^{2x} + 1 + \frac{1}{4e^{2x}}$

By taking $u = e^x + \frac{1}{2e^x}$, we see that:

$u^2 = \left( e^x + \frac{1}{2e^x} \right)^2 = e^{2x} + \frac{1}{4e^{2x}} + 1$

So we have:

$e^{2x} + 1 + \frac{e^{-2x}}{4} = \left( e^x + \frac{1}{2e^x} \right)^2$

Therefore, the integral simplifies to:

$L = 2 \int_{0}^{\ln 3} \sqrt{\left( e^x + \frac{1}{2e^x} \right)^2} e^x \, dx$

Since $\sqrt{u^2} = |u|$, and considering $e^x$ and $\frac{1}{2e^x}$ are positive within the range $0$ to $\ln 3$, we can drop the absolute value:

$L = 2 \int_{0}^{\ln 3} \left( e^x + \frac{1}{2e^x} \right) e^x \, dx$

Now, distribute $e^x$:

$L = 2 \int_{0}^{\ln 3} \left( e^{2x} + \frac{1}{2} \right) \, dx$

Separate the integrals:

$L = 2 \left( \int_{0}^{\ln 3} e^{2x} \, dx + \frac{1}{2} \int_{0}^{\ln 3} \, dx \right)$

Let's evaluate each integral separately.

For the first integral:

$\int_{0}^{\ln 3} e^{2x} \, dx$

Using the substitution $u = 2x$, $du = 2dx$, $dx = \frac{1}{2} du$:

$\int_{0}^{\ln 3} e^{2x} \, dx = \frac{1}{2} \int_{0}^{2 \ln 3} e^u \, du$

Evaluating the integral:

$\frac{1}{2} \left[ e^u \right]_{0}^{2 \ln 3} = \frac{1}{2} \left( e^{2 \ln 3} - 1 \right) = \frac{1}{2} \left( (e^{\ln 3})^2 - 1 \right) = \frac{1}{2} \left( 3^2 - 1 \right) = \frac{1}{2} \left( 9 - 1 \right) = \frac{1}{2} \cdot 8 = 4$

For the second integral:

$\frac{1}{2} \int_{0}^{\ln 3} \, dx = \frac{1}{2} \left[ x \right]_{0}^{\ln 3} = \frac{1}{2} (\ln 3 - 0) = \frac{1}{2} \ln 3$

Putting it all together:

$L = 2 \left( 4 + \frac{1}{2} \ln 3 \right) = 8 + \ln 3$

Therefore, the value of the integral is:

$L = 8 + \ln 3$

Would you like more details or have any questions about this solution?

Here are some related questions for further exploration:

1. How would you solve a similar integral with different limits of integration?
2. How does the integral change if the integrand is modified slightly?
3. What are some common techniques for simplifying integrals involving exponentials?
4. How can substitution methods be used to evaluate more complex integrals?
5. What is the significance of the natural logarithm in integral calculus?
6. How can integrals be evaluated numerically if an analytical solution is difficult to obtain?
7. What are some real-world applications of integrals involving exponential functions?
8. How does the presence of a square root in the integrand affect the approach to solving the integral?

Tip: When faced with complex integrals, breaking down the integrand into simpler parts can often reveal substitutions or simplifications that make the integral more manageable.