`I= int_6^16 (lnx^2)/(lnx^2+ln(x-22)^2)dx ?`

Answer

To solve the integral \( I = \int_{6}^{16} \frac{\ln x^2}{\ln x^2 + \ln(x-22)^2} \, dx \), we will utilize the properties of definite integrals, specifically a property known as King's rule.

### Step-by-step Solution:

1. **Set up the integral**:
\[ I = \int_{6}^{16} \frac{\ln x^2}{\ln x^2 + \ln(x-22)^2} \, dx \]

2. **Apply King's Rule**:
According to King's rule, we can express the integral as:
\[ I = \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
Here, \( a = 6 \) and \( b = 16 \). Thus, \( a + b = 22 \). We replace \( x \) with \( 22 - x \):
\[ I = \int_{6}^{16} \frac{\ln(22-x)^2}{\ln(22-x)^2 + \ln x^2} \, dx \]

3. **Rewrite the integral**:
Now we have two expressions for \( I \):
\[ I = \int_{6}^{16} \frac{\ln x^2}{\ln x^2 + \ln(x-22)^2} \, dx \quad \text{(Equation 1)} \]
\[ I = \int_{6}^{16} \frac{\ln(22-x)^2}{\ln(22-x)^2 + \ln x^2} \, dx \quad \text{(Equation 2)} \]

4. **Add the two equations**:
Adding Equation 1 and Equation 2:
\[ 2I = \int_{6}^{16} \left( \frac{\ln x^2}{\ln x^2 + \ln(x-22)^2} + \frac{\ln(22-x)^2}{\ln(22-x)^2 + \ln x^2} \right) dx \]

5. **Simplify the integrand**:
The sum of the two fractions simplifies to:
\[ \frac{\ln x^2 + \ln(22-x)^2}{\ln x^2 + \ln(22-x)^2} = 1 \]
Thus, we have:
\[ 2I = \int_{6}^{16} 1 \, dx \]

6. **Evaluate the integral**:
The integral of 1 from 6 to 16 is:
\[ \int_{6}^{16} 1 \, dx = 16 - 6 = 10 \]
Therefore, we have:
\[ 2I = 10 \]

7. **Solve for \( I \)**:
Dividing both sides by 2 gives:
\[ I = \frac{10}{2} = 5 \]

### Final Answer:
\[ I = 5 \]