Key Concepts and Formulas

• Property of Definite Integrals (King's Rule): For a continuous function $f(x)$ on the interval $[a, b]$, $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$.
• Logarithm Properties: $\log A^B = B \log A$.
• Perfect Square Trinomial: $a^2 - 2ab + b^2 = (a-b)^2$.

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Step-by-Step Solution:

Step 1: Identify and Simplify the Integrand Let the given integral be $I$. $I = \int\limits_2^4 {{{\log \,{x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}dx}$ The term $36 - 12x + x^2$ in the denominator is a perfect square trinomial, which can be written as $(x-6)^2$ or $(6-x)^2$. $(6-x)^2 = 36 - 12x + x^2$ Substituting this back into the integral, we get: $I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}dx} \quad \ldots(1)$ Why this step? Simplifying the quadratic expression reveals a potential symmetry in the integrand, which is a strong indicator that King's Rule might be applicable and effective.

Step 2: Apply King's Rule We apply the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a=2$ and $b=4$, so $a+b-x = 2+4-x = 6-x$. We replace every $x$ in the integrand with $(6-x)$.

The numerator becomes: $\log (6-x)^2$. The terms in the denominator become: $\log x^2$ becomes $\log (6-x)^2$. $\log (6-x)^2$ becomes $\log (6-(6-x))^2 = \log (6-6+x)^2 = \log x^2$.

So, the integral $I$ transforms into a new integral, let's call it $I'$: $I' = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}dx}$ Since the integrand is just transformed by replacing $x$ with $a+b-x$, the value of the integral remains the same. Thus, $I' = I$. $I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}dx} \quad \ldots(2)$ Why this step? King's Rule is applied to exploit the symmetry in the integrand. By substituting $a+b-x$, we aim to create a situation where adding the original and transformed integrals leads to significant simplification. Notice how the arguments of the logarithm functions in the numerator and denominator have swapped places.

Step 3: Add the Original and Transformed Integrals We add equation (1) and equation (2): $I + I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}dx} + \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}dx}$ Since the limits of integration are the same, we can combine them: $2I = \int\limits_2^4 {{{\log {x^2} + \log {{\left( {6 - x} \right)}^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}dx}$ The numerator and the denominator of the integrand are identical. Therefore, the integrand simplifies to $1$. $2I = \int\limits_2^4 {1\,dx}$ Why this step? Adding the two forms of the integral is the standard procedure after applying King's Rule. This step is designed to make the numerator and denominator identical, leading to a cancellation and a much simpler integrand.

Step 4: Evaluate the Simplified Integral Now we evaluate the integral of $1$ with respect to $x$: $2I = \left[ x \right]_2^4$ Applying the limits of integration: $2I = 4 - 2$ $2I = 2$ Solving for $I$: $I = \frac{2}{2}$ $I = 1$ Why this step? This is the final calculation to find the value of the original integral. Once the integrand simplifies to a constant, the integration process becomes straightforward.

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Common Mistakes & Tips:

• Recognizing Perfect Squares: Always be on the lookout for quadratic expressions that can be factored into perfect squares, as this often simplifies the integrand considerably.
• Applying King's Rule Correctly: Ensure that every instance of $x$ in the integrand is replaced by $a+b-x$. Double-check the transformation of each term, especially in complex expressions.
• Logarithm Properties: While $\log x^2 = 2 \log x$ is true, in this specific problem, applying it would lead to $2 \log x / (2 \log x + 2 \log (6-x))$, where the factor of 2 cancels out from the numerator and denominator. So, it's not strictly necessary for simplification here but is a useful property to recall.

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Summary

The given integral was evaluated using a combination of algebraic simplification and a key property of definite integrals, known as King's Rule. First, the quadratic term in the denominator was recognized as a perfect square, $(6-x)^2$. Then, King's Rule ($\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$) was applied with $a=2$ and $b=4$, transforming the integral. Adding the original and the transformed integrals resulted in an integrand of $1$, which was then easily integrated. This process led to the value of the integral being $1$.

The final answer is $\boxed{1}$.