Answer:
Option C
Explanation:
Central Idea.
Apply the property $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ and then add.
Let $I=\int_{2}^{4} \frac{\log x^{2}}{\log x^{2}+\log (36-12x+x^{2})}dx$
$I= \int_{2}^{4} \frac{2\log x}{2\log x+\log(6-x)^{2}}dx$
$= \int_{2}^{4} \frac{2\log x dx}{2[\log x+\log(6-x)^{}]}$
$\Rightarrow I=\int_{2}^{4} \frac{\log x dx}{[\log x+\log(6-x)^{}]}$ ......(i)
$\Rightarrow I=\int_{2}^{4} \frac{\log (6-x) }{\log (6-x)+\log x}dx$ .......(ii)
$\left[\therefore \int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx\right]$
On adding Eqs .(i) and (ii) , we get
$2I= \int_{2}^{4}\frac{\log x +\log (6-x)}{\log x+\log(6-x)}dx$
$2I= \int_{2}^{4}dx=\left[ x\right]_2^4$
$\Rightarrow 2I=2\Rightarrow I=1$
