The integral $\int\limits_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x$, is equal to

1

Let $I=\int\limits_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} d x$. Then,

$I=\int\limits_2^4 \frac{\log x^2}{\log x^2+\log (6-x)^2} d x$

Clearly, it is of the form $\int\limits_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x$, where $f(x)=\log x^2, a=2$ and $b=4$

We know that $\int\limits_a^b \frac{f(x)}{f(x)+f(a+b-x)} d x=\frac{b-a}{2}$

∴   $I=\frac{4-2}{2}=1$