The integral $$\int_2^4 \frac{\log x^2}{\log x^2 + \log(6-x)^2} dx$$ is equal to

The given integral is $$\int_2^4 \frac{\log x^2}{\log x^2 + \log(6-x)^2} dx$$. First, simplify the logarithmic expressions. For $$x$$ in the interval $$[2, 4]$$, $$x > 0$$ and $$6 - x > 0$$, so $$\log x^2 = \log (x^2) = 2 \log x$$ and $$\log (6-x)^2 = \log ((6-x)^2) = 2 \log (6-x)$$. Substitute these into the integral: $$\int_2^4 \frac{2 \log x}{2 \log x + 2 \log (6-x)} dx$$ Factor out the 2 in the denominator: $$\int_2^4 \frac{2 \log x}{2 (\log x + \log (6-x))} dx = \int_2^4 \frac{\log x}{\log x + \log (6-x)} dx$$ Use the logarithm property $$\log a + \log b = \log (a b)$$ to combine the denominator: $$\log x + \log (6-x) = \log [x(6-x)]$$ So the integral becomes: $$I = \int_2^4 \frac{\log x}{\log [x(6-x)]} dx$$ Apply the property of definite integrals: $$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$. Here, $$a = 2$$ and $$b = 4$$, so $$a + b = 6$$. Replace $$x$$ with $$6 - x$$: $$I = \int_2^4 \frac{\log (6-x)}{\log [(6-x)(6-(6-x))]} dx = \int_2^4 \frac{\log (6-x)}{\log [(6-x) \cdot x]} dx = \int_2^4 \frac{\log (6-x)}{\log [x(6-x)]} dx$$ Add the original expression for $$I$$ and the expression after substitution: $$I + I = \int_2^4 \frac{\log x}{\log [x(6-x)]} dx + \int_2^4 \frac{\log (6-x)}{\log [x(6-x)]} dx$$ Combine the integrals: $$2I = \int_2^4 \left( \frac{\log x}{\log [x(6-x)]} + \frac{\log (6-x)}{\log [x(6-x)]} \right) dx = \int_2^4 \frac{\log x + \log (6-x)}{\log [x(6-x)]} dx$$ The numerator $$\log x + \log (6-x) = \log [x(6-x)]$$, so: $$2I = \int_2^4 \frac{\log [x(6-x)]}{\log [x(6-x)]} dx = \int_2^4 1 dx$$ Evaluate the integral: $$\int_2^4 1 dx = [x]_2^4 = 4 - 2 = 2$$ So: $$2I = 2 \implies I = 1$$ The value of the integral is 1. Hence, the correct answer is Option D.