To solve the integral
\[
I = \int_{\ln 3}^{\ln 4} \frac{e^x \sqrt{e^x - 3}}{e^x - 2} \, dx,
\]
we will use a substitution method.
### Step 1: Substitution
Let \( t = e^x - 2 \). Then, we differentiate to find \( dx \):
\[
dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x}.
\]
Since \( e^x = t + 2 \), we can substitute \( dx \) as:
\[
dx = \frac{dt}{t + 2}.
\]
### Step 2: Change the limits of integration
Now we need to change the limits of integration. When \( x = \ln 3 \):
\[
t = e^{\ln 3} - 2 = 3 - 2 = 1.
\]
When \( x = \ln 4 \):
\[
t = e^{\ln 4} - 2 = 4 - 2 = 2.
\]
Thus, the limits change from \( x = \ln 3 \) to \( x = \ln 4 \) into \( t = 1 \) to \( t = 2 \).
### Step 3: Substitute into the integral
Now substitute \( t \) into the integral:
\[
I = \int_{1}^{2} \frac{(t + 2) \sqrt{(t + 2) - 3}}{t} \cdot \frac{dt}{t + 2}.
\]
This simplifies to:
\[
I = \int_{1}^{2} \frac{\sqrt{t - 1}}{t} \, dt.
\]
### Step 4: Further substitution
Let \( u = \sqrt{t - 1} \), then \( t = u^2 + 1 \) and \( dt = 2u \, du \). The limits change as follows:
- When \( t = 1 \), \( u = 0 \).
- When \( t = 2 \), \( u = 1 \).
Thus, the integral becomes:
\[
I = \int_{0}^{1} \frac{u}{u^2 + 1} \cdot 2u \, du = 2 \int_{0}^{1} \frac{u^2}{u^2 + 1} \, du.
\]
### Step 5: Simplifying the integral
We can split the integral:
\[
I = 2 \left( \int_{0}^{1} 1 \, du - \int_{0}^{1} \frac{1}{u^2 + 1} \, du \right).
\]
The first integral evaluates to:
\[
\int_{0}^{1} 1 \, du = 1.
\]
The second integral is:
\[
\int_{0}^{1} \frac{1}{u^2 + 1} \, du = \frac{\pi}{4}.
\]
Thus, we have:
\[
I = 2 \left( 1 - \frac{\pi}{4} \right) = 2 - \frac{\pi}{2}.
\]
### Final Result
The value of the integral is:
\[
\boxed{2 - \frac{\pi}{2}}.
\]
