int_1^3 frac{log x^2}{log left(16 x^2-8 x^3+x | vedclass

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(A) Let $I = \int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} dx$. Note that $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(4-x)^2$. Thus,

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$1$

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(A) Let $I = \int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} dx$. Note that $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(4-x)^2$. Thus,the denominator is $\log(x^2(4-x)^2) = \log x^2 + \log(4-x)^2$. So,$I = \int_1^3 \frac{\log x^2}{\log x^2 + \log(4-x)^2} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 1+3 = 4$. Replacing $x$ with $(4-x)$,we get $I = \int_1^3 \frac{\log(4-x)^2}{\log(4-x)^2 + \log x^2} dx$. Adding the two expressions for $I$: $2I = \int_1^3 \frac{\log x^2 + \log(4-x)^2}{\log x^2 + \log(4-x)^2} dx = \int_1^3 1 dx = [x]_1^3 = 3-1 = 2$. Therefore,$I = 1$.

$\int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} d x=\ldots$

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