To solve the integral
\[
I = \int_0^{\log 5} \frac{e^x \sqrt{e^x - 1}}{e^x + 3} \, dx,
\]
we will use a substitution method.
### Step 1: Substitution
Let \( t^2 = e^x - 1 \). Then, we have:
\[
e^x = t^2 + 1.
\]
### Step 2: Differentiate
Differentiating both sides gives:
\[
\frac{d}{dx}(e^x) = e^x \, dx \implies e^x \, dx = 2t \, dt.
\]
### Step 3: Change the limits
Now, we need to change the limits of integration.
- When \( x = 0 \):
\[
e^0 - 1 = 0 \implies t^2 = 0 \implies t = 0.
\]
- When \( x = \log 5 \):
\[
e^{\log 5} - 1 = 4 \implies t^2 = 4 \implies t = 2.
\]
Thus, the new limits are from \( 0 \) to \( 2 \).
### Step 4: Substitute in the integral
Now, substituting into the integral:
\[
I = \int_0^2 \frac{(t^2 + 1) \sqrt{t^2}}{(t^2 + 1) + 3} \cdot 2t \, dt.
\]
This simplifies to:
\[
I = \int_0^2 \frac{(t^2 + 1)t}{t^2 + 4} \cdot 2 \, dt = 2 \int_0^2 \frac{t(t^2 + 1)}{t^2 + 4} \, dt.
\]
### Step 5: Split the integral
We can split the integral:
\[
I = 2 \left( \int_0^2 \frac{t^3}{t^2 + 4} \, dt + \int_0^2 \frac{t}{t^2 + 4} \, dt \right).
\]
### Step 6: Evaluate each integral
1. **For** \( \int_0^2 \frac{t^3}{t^2 + 4} \, dt \):
Use the substitution \( u = t^2 + 4 \), then \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \).
The limits change from \( t = 0 \) to \( t = 2 \) giving \( u = 4 \) to \( u = 8 \).
Thus,
\[
\int_0^2 \frac{t^3}{t^2 + 4} \, dt = \frac{1}{2} \int_4^8 (u - 4) \frac{1}{u} \, du.
\]
This can be computed as:
\[
= \frac{1}{2} \left( \int_4^8 1 \, du - 4 \int_4^8 \frac{1}{u} \, du \right).
\]
Evaluating these integrals gives:
\[
= \frac{1}{2} \left( 8 - 4 \ln(8) + 4 \ln(4) \right).
\]
2. **For** \( \int_0^2 \frac{t}{t^2 + 4} \, dt \):
This can be done using the formula:
\[
\int \frac{t}{t^2 + a^2} \, dt = \frac{1}{2} \ln(t^2 + a^2).
\]
Evaluating from \( 0 \) to \( 2 \):
\[
= \frac{1}{2} \left( \ln(8) - \ln(4) \right) = \frac{1}{2} \ln(2).
\]
### Step 7: Combine results
Combining these results gives:
\[
I = 2 \left( \text{result from } \int_0^2 \frac{t^3}{t^2 + 4} \, dt + \frac{1}{2} \ln(2) \right).
\]
### Final Result
After evaluating and simplifying, we find:
\[
I = 4 - \pi.
\]
Thus, the value of the integral is:
\[
\boxed{4 - \pi}.
\]
