`int_(2-ln3)^(3+ln3)(ln(4+x))/(ln(4+x)+ln(9-x))dx` is equal to :

Answer

To solve the integral

\[ I = \int_{2 - \ln 3}^{3 + \ln 3} \frac{\ln(4 + x)}{\ln(4 + x) + \ln(9 - x)} \, dx, \]

we can use the property of definite integrals which states that:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

### Step 1: Identify the limits and the function
Here, we have \( a = 2 - \ln 3 \) and \( b = 3 + \ln 3 \). The sum \( a + b \) is:

\[ (2 - \ln 3) + (3 + \ln 3) = 5. \]

### Step 2: Apply the property
Now, we apply the property to our integral:

\[ I = \int_{2 - \ln 3}^{3 + \ln 3} \frac{\ln(4 + (5 - x))}{\ln(4 + (5 - x)) + \ln(9 - (5 - x))} \, dx. \]

### Step 3: Simplify the function
Now simplify the expression inside the integral:

\[ 4 + (5 - x) = 9 - x, \]
\[ 9 - (5 - x) = 4 + x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{2 - \ln 3}^{3 + \ln 3} \frac{\ln(9 - x)}{\ln(9 - x) + \ln(4 + x)} \, dx. \]

### Step 4: Set up the two equations
Now we have two expressions for \( I \):

1. \( I = \int_{2 - \ln 3}^{3 + \ln 3} \frac{\ln(4 + x)}{\ln(4 + x) + \ln(9 - x)} \, dx \) (original)
2. \( I = \int_{2 - \ln 3}^{3 + \ln 3} \frac{\ln(9 - x)}{\ln(9 - x) + \ln(4 + x)} \, dx \) (transformed)

### Step 5: Add the two equations
Adding these two equations gives:

\[ 2I = \int_{2 - \ln 3}^{3 + \ln 3} \left( \frac{\ln(4 + x)}{\ln(4 + x) + \ln(9 - x)} + \frac{\ln(9 - x)}{\ln(9 - x) + \ln(4 + x)} \right) dx. \]

### Step 6: Simplify the integrand
The sum of the fractions simplifies to:

\[ \frac{\ln(4 + x) + \ln(9 - x)}{\ln(4 + x) + \ln(9 - x)} = 1. \]

Thus, we have:

\[ 2I = \int_{2 - \ln 3}^{3 + \ln 3} 1 \, dx. \]

### Step 7: Evaluate the integral
Now we can evaluate the integral:

\[ \int_{2 - \ln 3}^{3 + \ln 3} 1 \, dx = (3 + \ln 3) - (2 - \ln 3) = 3 + \ln 3 - 2 + \ln 3 = 1 + 2\ln 3. \]

### Step 8: Solve for \( I \)
Now, we can solve for \( I \):

\[ 2I = 1 + 2\ln 3 \implies I = \frac{1 + 2\ln 3}{2}. \]

### Final Answer
Thus, the value of the integral is:

\[ I = \frac{1}{2} + \ln 3. \]