In the coordinate plane, find the area of the region bounded by the curve $ C: y\equal{}\frac{x\plus{}1}{x^2\plus{}1}$ and the line $ L: y\equal{}1$.
1. **Define the function and the line:**
   The curve \( C \) is given by the function \( f(x) = \frac{x+1}{x^2+1} \) and the line \( L \) is given by \( y = 1 \).

2. **Find the intersection points:**
   To find the points where the curve intersects the line \( y = 1 \), we solve the equation:
   \[
   \frac{x+1}{x^2+1} = 1
   \]
   Multiply both sides by \( x^2 + 1 \):
   \[
   x + 1 = x^2 + 1
   \]
   Rearrange the equation:
   \[
   x^2 - x = 0
   \]
   Factorize:
   \[
   x(x - 1) = 0
   \]
   So, the solutions are \( x = 0 \) and \( x = 1 \).

3. **Evaluate the function at the intersection points:**
   - At \( x = 0 \):
     \[
     f(0) = \frac{0 + 1}{0^2 + 1} = 1
     \]
   - At \( x = 1 \):
     \[
     f(1) = \frac{1 + 1}{1^2 + 1} = \frac{2}{2} = 1
     \]

4. **Determine the region of interest:**
   The region bounded by the curve \( C \) and the line \( L \) is between \( x = 0 \) and \( x = 1 \).

5. **Set up the integral to find the area:**
   The area \( A \) between the curve \( y = \frac{x+1}{x^2+1} \) and the line \( y = 1 \) from \( x = 0 \) to \( x = 1 \) is given by:
   \[
   A = \int_{0}^{1} \left(1 - \frac{x+1}{x^2+1}\right) \, dx
   \]

6. **Simplify the integrand:**
   \[
   1 - \frac{x+1}{x^2+1} = \frac{x^2 + 1 - (x + 1)}{x^2 + 1} = \frac{x^2 - x}{x^2 + 1}
   \]

7. **Evaluate the integral:**
   \[
   A = \int_{0}^{1} \frac{x^2 - x}{x^2 + 1} \, dx
   \]
   Split the integral:
   \[
   A = \int_{0}^{1} \frac{x^2}{x^2 + 1} \, dx - \int_{0}^{1} \frac{x}{x^2 + 1} \, dx
   \]

8. **Evaluate each part separately:**
   - For the first part, use the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \):
     \[
     \int_{0}^{1} \frac{x^2}{x^2 + 1} \, dx = \int_{0}^{1} \left(1 - \frac{1}{x^2 + 1}\right) \, dx = \int_{0}^{1} 1 \, dx - \int_{0}^{1} \frac{1}{x^2 + 1} \, dx
     \]
     \[
     = \left[ x \right]_{0}^{1} - \left[ \arctan(x) \right]_{0}^{1} = 1 - \frac{\pi}{4}
     \]

   - For the second part, use the substitution \( u = x^2 + 1 \), \( du = 2x \, dx \):
     \[
     \int_{0}^{1} \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du = \frac{1}{2} \left[ \ln(u) \right]_{1}^{2} = \frac{1}{2} (\ln(2) - \ln(1)) = \frac{\ln(2)}{2}
     \]

9. **Combine the results:**
   \[
   A = \left(1 - \frac{\pi}{4}\right) - \frac{\ln(2)}{2}
   \]

The final answer is \(\boxed{1 - \frac{\pi}{4} - \frac{\ln(2)}{2}}\)