The integral you've provided is:

$\int e^{2x} \sqrt{1 - e^{x}} \, dx$

To solve this integral, we can use substitution. Let's proceed with the following steps:

Step 1: Substitution

Let $u = 1 - e^{x}$. Then, the derivative of $u$ with respect to $x$ is:

$\frac{du}{dx} = -e^{x}$

This implies:

$du = -e^{x} \, dx \quad \text{or} \quad -du = e^{x} \, dx$

Step 2: Rewrite the Integral

Now, substitute $u = 1 - e^{x}$ and $e^{x} \, dx = -du$ into the integral:

$\int e^{2x} \sqrt{1 - e^{x}} \, dx = \int e^{x} \cdot e^{x} \sqrt{u} \, dx$

Substitute $e^{x} \, dx = -du$:

$= -\int e^{x} \cdot \sqrt{u} \, du$

Since $e^{x} = 1 - u$:

$= -\int (1 - u) \sqrt{u} \, du$

Step 3: Expand and Integrate

Now, distribute and integrate:

$= -\int \sqrt{u} \, du + \int u \sqrt{u} \, du$

These integrals are straightforward:

1. For $\int \sqrt{u} \, du$, recall that $\sqrt{u} = u^{1/2}$:

$\int u^{1/2} \, du = \frac{2}{3} u^{3/2}$

2. For $\int u \sqrt{u} \, du$, note that $u \sqrt{u} = u^{3/2}$:

$\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$

Step 4: Combine and Back-Substitute

Combining these, we get:

$-\left[\frac{2}{3} u^{3/2} - \frac{2}{5} u^{5/2}\right] + C$

Substituting back $u = 1 - e^{x}$:

$-\left[\frac{2}{3} (1 - e^{x})^{3/2} - \frac{2}{5} (1 - e^{x})^{5/2}\right] + C$

This is the final answer for the integral.

Would you like further details or explanations? Here are some related questions you might find interesting:

1. What is the method of substitution in integration?
2. How can we verify the result of an integral?
3. What are some common techniques for solving integrals involving square roots?
4. What role do differentials play in substitution methods?
5. How do you handle integrals involving exponential functions?

Tip: When dealing with integrals that seem complex, substitution can often simplify the expression significantly, making it easier to integrate.