The integral `int_(2)^(4)(logx^(2))/(logx^(2)+log(36-12x+x^(2))) dx` is equal to

Options

A. `2`
B. `4`
C. `1`
D. `6`

Explanation

To solve the integral

\[ I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log(36 - 12x + x^2)} \, dx, \]

we will follow these steps:

### Step 1: Simplify the expression inside the integral

First, observe that

\[ 36 - 12x + x^2 = (x - 6)^2. \]

Thus, we can rewrite the integral as:

\[ I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log((x - 6)^2)} \, dx. \]

Using the property of logarithms, we can simplify:

\[ \log((x - 6)^2) = 2 \log(x - 6). \]

So, we have:

\[ I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + 2 \log(x - 6)} \, dx. \]

### Step 2: Use the property of definite integrals

We will use the property that states:

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

Here, \( a = 2 \) and \( b = 4 \), so:

\[ I = \int_{2}^{4} \frac{\log((2 + 4 - x)^2)}{\log((2 + 4 - x)^2) + \log((2 + 4 - x - 6)^2)} \, dx. \]

Calculating \( 2 + 4 - x \):

\[ 2 + 4 - x = 6 - x. \]

Thus, we can rewrite the integral as:

\[ I = \int_{2}^{4} \frac{\log((6 - x)^2)}{\log((6 - x)^2) + \log((6 - x - 6)^2)} \, dx = \int_{2}^{4} \frac{\log((6 - x)^2)}{\log((6 - x)^2) + \log((x - 6)^2)} \, dx. \]

### Step 3: Combine the two expressions for \( I \)

Now we have two expressions for \( I \):

1. \( I = \int_{2}^{4} \frac{\log x^2}{\log x^2 + 2 \log(x - 6)} \, dx \)
2. \( I = \int_{2}^{4} \frac{\log((6 - x)^2)}{\log((6 - x)^2) + 2 \log(x - 6)} \, dx \)

Adding these two equations:

\[ 2I = \int_{2}^{4} \left( \frac{\log x^2 + \log((6 - x)^2)}{\log x^2 + 2 \log(x - 6)} \right) \, dx. \]

### Step 4: Simplify the numerator

Using the property of logarithms:

\[ \log x^2 + \log((6 - x)^2) = \log(x^2(6 - x)^2) = \log(6^2 - 12x + x^2). \]

### Step 5: Evaluate the integral

Now, we can simplify the expression:

\[ 2I = \int_{2}^{4} \frac{\log(6^2 - 12x + x^2)}{\log(6^2 - 12x + x^2)} \, dx = \int_{2}^{4} 1 \, dx = 4 - 2 = 2. \]

Thus, we have:

\[ 2I = 2 \implies I = 1. \]

### Final Answer

The value of the integral is

\[ \boxed{1}. \]