The integral ∫24 log⁡x2log⁡x2+log⁡36−12x+x2dx is equal to

Let $I = \int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$ , Then,

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

Clearly, it is of the form $\int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x,$ where

$f (x) = \log x^{2}, a = 2 and b = 4$ .

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

$∴ I = \frac{4 - 2}{2} = 1$ .

The integral $\int_{2}^{4} \frac{\log x^{2}}{\log x^{2} + \log (36 - 12 x + x^{2})} d x$ is equal to