1. Rewrite the denominator

We need to evaluate

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

Notice that

$$36-12x+x^2=x^2-12x+36=(x-6)^2=(6-x)^2.$$

So

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

Using $\log a+\log b=\log(ab)$,

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

2. Use symmetry with the substitution $x\mapsto 6-x$

Let

$$f(x)=\frac{\log x^2}{\log x^2+\log(6-x)^2}.$$

Then

$$I=\int_2^4 f(x)\,dx.$$

Now apply the substitution

$$x=6-t \quad \Rightarrow \quad dx=-dt.$$

When $x=2$, $t=4$; when $x=4$, $t=2$. Hence

$$I=\int_4^2 f(6-t)(-dt)=\int_2^4 f(6-t)\,dt.$$

Renaming $t$ as $x$,

$$I=\int_2^4 f(6-x)\,dx.$$

Now compute

$$f(6-x)=\frac{\log(6-x)^2}{\log(6-x)^2+\log x^2}.$$

Therefore,

$$f(x)+f(6-x)=\frac{\log x^2}{\log x^2+\log(6-x)^2}+\frac{\log(6-x)^2}{\log x^2+\log(6-x)^2}=1.$$

3. Add the two expressions for $I$

Since

$$I=\int_2^4 f(x)\,dx$$

and

$$I=\int_2^4 f(6-x)\,dx,$$

adding,

$$2I=\int_2^4 \big(f(x)+f(6-x)\big)\,dx=\int_2^4 1\,dx=4-2=2.$$

So

$$I=1.$$

4. Check options

Thus the value of the integral is

$$\boxed{1}.$$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.