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

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(A) माना $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. यहाँ $36-12x+x^2 = (6-x)^2$ है। अतः,$I = \int_2^4 \frac{\log x^2}{\log x^2 +

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$1$

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(A) माना $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. यहाँ $36-12x+x^2 = (6-x)^2$ है। अतः,$I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (6-x)^2} dx$. गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करने पर,जहाँ $a+b = 2+4 = 6$. $x$ को $6-x$ से प्रतिस्थापित करने पर: $I = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log (6-(6-x))^2} dx = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log x^2} dx$. $I$ के दोनों समीकरणों को जोड़ने पर: $2I = \int_2^4 \frac{\log x^2 + \log (6-x)^2}{\log x^2 + \log (6-x)^2} dx = \int_2^4 1 dx$. $2I = [x]_2^4 = 4-2 = 2$. अतः,$I = 1$.

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ का मान ज्ञात कीजिए।

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