int_2^4 frac{log x^2}{log x^2+log (36-12x+x^2 | vedclass

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(A) Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. Note that $36-12x+x^2 = (6-x)^2$. So,$I = \int_2^4 \frac{\log x^2}{\log x^2 +

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(A) Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. Note that $36-12x+x^2 = (6-x)^2$. So,$I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (6-x)^2} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 2+4 = 6$. Replacing $x$ with $6-x$: $I = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log (6-(6-x))^2} dx = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log x^2} dx$. Adding the two expressions for $I$: $2I = \int_2^4 \frac{\log x^2 + \log (6-x)^2}{\log x^2 + \log (6-x)^2} dx = \int_2^4 1 dx$. $2I = [x]_2^4 = 4-2 = 2$. Therefore,$I = 1$.

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ is equal to

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