{Given the integral } \[\int_{5}^{9} \frac{\log 3x^2}{\log 3x^2 + \log (588 - 84x + 3x^2)} \, dx \quad \cdots (i)\]The second term in the denominator can be rewritten as:
\[\log (588 - 84x + 3x^2) = \log (3(196 - 28x + x^2)) = \log (3(14-x)^2)\]{Applying the property of definite integrals, } \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx, { yields:}\[I = \int_{5}^{9} \frac{\log (3(14-x)^2)}{\log (3(14-x)^2) + \log (3x^2)} \, dx\]\[I = \int_{5}^{9} \frac{\log 3 + 2\log (14-x)}{\log 3 + 2\log (14-x) + \log 3 + 2\log x} \, dx\]\[I = \int_{5}^{9} \frac{\log 3 + 2\log (14-x)}{2\log 3 + 2\log (14-x) + 2\log x} \, dx\]\[I = \int_{5}^{9} \frac{\log 3 + 2\log (14-x)}{2(\log 3 + \log (14-x) + \log x)} \, dx\]\[I = \int_{5}^{9} \frac{\log 3 + 2\log (14-x)}{2\log (3x(14-x))} \, dx \quad \cdots (ii)\]{Summing equations (i) and (ii):}\[2I = \int_{5}^{9} \frac{\log 3x^2 + \log (3(14-x)^2)}{\log 3x^2 + \log (3(14-x)^2)} \, dx\]\[2I = \int_{5}^{9} 1 \, dx\]\[2I = [x]_5^9\]\[2I = 9 - 5 = 4\]I = 2