$\int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} \mathrm{d} x$ is equal to

Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log(36-12x+x^2)}dx$.

Observe that $36 - 12x + x^2 = (6-x)^2$, so the integral becomes:

$I = \int_2^4 \frac{\log x^2}{\log x^2 + \log(6-x)^2}dx$

Using the substitution property $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$ with $a=2$, $b=4$:

$I = \int_2^4 \frac{\log(6-x)^2}{\log(6-x)^2 + \log x^2}dx$

Adding these two expressions for $I$:

$2I = \int_2^4 \frac{\log x^2 + \log(6-x)^2}{\log x^2 + \log(6-x)^2}dx = \int_2^4 1dx$

Evaluating the integral yields $2I = [x]_2^4 = 2$, hence $I = 1$.

Final answer: $\boxed{1}$