PLAN Apply the property
$\int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(a+b-x) d x$ and then add.
Let $=\int\limits_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log \left(36-12 x+x^{2}\right)} d x $
$=\int\limits_{2}^{4} \frac{2 \log x}{2 \log x+\log (6-x)^{2}} d x=\int\limits_{2}^{4} \frac{2 \log x d x}{2[\log x+\log (6-x)]} $
$\Rightarrow I =\int\limits_{2}^{4} \frac{\log x d x}{[\log x+\log (6-x)]} ....$(i)
$\Rightarrow I =\int\limits_{2}^{4} \frac{\log (6-x)}{\log (6-x)+\log x} d x ....$(ii)
${\left[\because \int\limits_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(a+b-x) d x\right] }$
On adding Eqs. (i) and (ii), we get
$ 2 I=\int\limits_{2}^{4} \frac{\log x+\log (6-x)}{\log x+\log (6-x)} d x $
$\Rightarrow 2 I=\int\limits_{2}^{4} d x=[x]_{2}^{4} \Rightarrow 2 I=2 $
$\Rightarrow 2 I=2 \Rightarrow I=1$