int_5^9 frac{log_3 x^2}{log_3 x^2 + log_3(588 | vedclass

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(A) Let $I = \int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx$. Using the property $\log_3(588 - 84x + 3x^2) = \log_3(3(196 - 28x

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$2$

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(A) Let $I = \int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx$. Using the property $\log_3(588 - 84x + 3x^2) = \log_3(3(196 - 28x + x^2)) = \log_3(3(14 - x)^2)$,we have: $I = \int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(3(14 - x)^2)} dx \dots (1)$. Using the property $\int_a^b f(x) dx = \int_a^b f(a + b - x) dx$,where $a=5$ and $b=9$,$a+b-x = 14-x$: $I = \int_5^9 \frac{\log_3(14 - x)^2}{\log_3(14 - x)^2 + \log_3(3(14 - (14 - x))^2)} dx$. $I = \int_5^9 \frac{\log_3(14 - x)^2}{\log_3(14 - x)^2 + \log_3(3x^2)} dx \dots (2)$. Adding $(1)$ and $(2)$: $2I = \int_5^9 \frac{\log_3 x^2 + \log_3(3(14 - x)^2)}{\log_3 x^2 + \log_3(3(14 - x)^2)} dx$. $2I = \int_5^9 1 dx = [x]_5^9 = 9 - 5 = 4$. $I = \frac{4}{2} = 2$.

$\int_5^9 \frac{\log_3 x^2}{\log_3 x^2 + \log_3(588 - 84x + 3x^2)} dx =$

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