To solve the problem, we need to evaluate the integral \( I(m) = \int_0^\pi \ln(1 - 2m \cos x + m^2) \, dx \) at \( m = 1 \).
### Step-by-Step Solution:
1. **Substituting \( m = 1 \)**:
\[
I(1) = \int_0^\pi \ln(1 - 2(1) \cos x + 1^2) \, dx
\]
Simplifying the expression inside the logarithm:
\[
I(1) = \int_0^\pi \ln(1 - 2 \cos x + 1) \, dx = \int_0^\pi \ln(2 - 2 \cos x) \, dx
\]
2. **Factoring out the constant**:
We can factor out the 2 from the logarithm:
\[
I(1) = \int_0^\pi \ln(2(1 - \cos x)) \, dx = \int_0^\pi \ln(2) \, dx + \int_0^\pi \ln(1 - \cos x) \, dx
\]
Since \( \ln(2) \) is a constant, we can compute:
\[
\int_0^\pi \ln(2) \, dx = \ln(2) \cdot \pi
\]
3. **Evaluating \( \int_0^\pi \ln(1 - \cos x) \, dx \)**:
We can use the identity \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \):
\[
\int_0^\pi \ln(1 - \cos x) \, dx = \int_0^\pi \ln(2 \sin^2\left(\frac{x}{2}\right)) \, dx
\]
This can be split into two integrals:
\[
= \int_0^\pi \ln(2) \, dx + \int_0^\pi \ln(\sin^2\left(\frac{x}{2}\right)) \, dx
\]
The first part is:
\[
\int_0^\pi \ln(2) \, dx = \ln(2) \cdot \pi
\]
The second part can be simplified:
\[
= 2 \int_0^\pi \ln(\sin\left(\frac{x}{2}\right)) \, dx
\]
Using the known result \( \int_0^\pi \ln(\sin x) \, dx = -\pi \ln(2) \), we find:
\[
\int_0^\pi \ln(\sin\left(\frac{x}{2}\right)) \, dx = -\pi \ln(2) \quad \text{(as \( \sin\left(\frac{x}{2}\right) \) is half the angle)}
\]
4. **Combining the results**:
Thus, we have:
\[
\int_0^\pi \ln(1 - \cos x) \, dx = \pi \ln(2) - \pi \ln(2) = 0
\]
5. **Final Result**:
Therefore, combining everything:
\[
I(1) = \ln(2) \cdot \pi + 0 = \pi \ln(2)
\]
### Conclusion:
The final answer is:
\[
\boxed{\pi \ln(2)}
\]
