To evaluate the integral
\[
I = \int_{4}^{6} \frac{\lfloor x^2 \rfloor}{\lfloor x^2 - 20x + 100 \rfloor + \lfloor x^2 \rfloor} \, dx,
\]
we will follow these steps:
### Step 1: Simplify the expression in the integral
First, we simplify the expression in the denominator:
\[
x^2 - 20x + 100 = (x - 10)^2.
\]
Thus, we can rewrite the integral as:
\[
I = \int_{4}^{6} \frac{\lfloor x^2 \rfloor}{\lfloor (x - 10)^2 \rfloor + \lfloor x^2 \rfloor} \, dx.
\]
### Step 2: Evaluate the greatest integer function
Next, we need to evaluate the values of \(\lfloor x^2 \rfloor\) and \(\lfloor (x - 10)^2 \rfloor\) for \(x\) in the interval \([4, 6]\).
- For \(x = 4\):
\[
\lfloor 4^2 \rfloor = \lfloor 16 \rfloor = 16,
\]
\[
\lfloor (4 - 10)^2 \rfloor = \lfloor (-6)^2 \rfloor = \lfloor 36 \rfloor = 36.
\]
- For \(x = 5\):
\[
\lfloor 5^2 \rfloor = \lfloor 25 \rfloor = 25,
\]
\[
\lfloor (5 - 10)^2 \rfloor = \lfloor (-5)^2 \rfloor = \lfloor 25 \rfloor = 25.
\]
- For \(x = 6\):
\[
\lfloor 6^2 \rfloor = \lfloor 36 \rfloor = 36,
\]
\[
\lfloor (6 - 10)^2 \rfloor = \lfloor (-4)^2 \rfloor = \lfloor 16 \rfloor = 16.
\]
### Step 3: Determine the ranges for \(\lfloor x^2 \rfloor\) and \(\lfloor (x - 10)^2 \rfloor\)
From the calculations:
- For \(x \in [4, 5)\):
\(\lfloor x^2 \rfloor = 16\) and \(\lfloor (x - 10)^2 \rfloor = 36\).
- For \(x \in [5, 6)\):
\(\lfloor x^2 \rfloor = 25\) and \(\lfloor (x - 10)^2 \rfloor = 25\).
### Step 4: Break the integral into two parts
We can now break the integral into two parts:
\[
I = \int_{4}^{5} \frac{16}{36 + 16} \, dx + \int_{5}^{6} \frac{25}{25 + 25} \, dx.
\]
### Step 5: Evaluate each part
1. For the first integral:
\[
I_1 = \int_{4}^{5} \frac{16}{52} \, dx = \int_{4}^{5} \frac{4}{13} \, dx = \frac{4}{13} \cdot (5 - 4) = \frac{4}{13}.
\]
2. For the second integral:
\[
I_2 = \int_{5}^{6} \frac{25}{50} \, dx = \int_{5}^{6} \frac{1}{2} \, dx = \frac{1}{2} \cdot (6 - 5) = \frac{1}{2}.
\]
### Step 6: Combine the results
Now, we combine both results:
\[
I = I_1 + I_2 = \frac{4}{13} + \frac{1}{2}.
\]
To add these fractions, find a common denominator:
\[
\frac{1}{2} = \frac{13}{26} \quad \text{and} \quad \frac{4}{13} = \frac{8}{26}.
\]
Thus,
\[
I = \frac{8}{26} + \frac{13}{26} = \frac{21}{26}.
\]
### Final Answer
The value of the integral is
\[
\boxed{\frac{21}{26}}.
\]
