$I =\int\limits_{2}^{4} \frac{\log x ^{2}}{\log x ^{2}+\log ( x -6)^{2}} dx \ldots .(1)$
using $\int\limits_{a}^{b} f ( x ) d x =\int\limits_{ a }^{ b } f ( a + b - x ) d x$
$I =\int\limits_{2}^{4} \frac{\log (6- x )^{2}}{\log (6- x )^{2}+\log x ^{2}} dx \ldots(2)$
$(1)+(2)$ gives
$2 I =\int\limits_{2}^{4} 1 \,dx =2$
$I =1$