Key Concepts and Formulas

• Logarithm Properties: $\log_e a^n = n \log_e a$, $\log_e (ab) = \log_e a + \log_e b$, $\log_e a - \log_e b = \log_e (a/b)$.
• Perfect Square Formula: $(a-b)^2 = a^2 - 2ab + b^2$.
• King's Rule of Definite Integration: $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx$. This property is particularly useful when the integrand exhibits symmetry or when the sum $f(x) + f(a+b-x)$ simplifies.

Step-by-Step Solution

Step 1: Simplify the Integrand

Let the given integral be $I$. $I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}$ First, let's simplify the quadratic expression in the denominator: $x^2 - 44x + 484$. We recognize this as a perfect square trinomial: $x^2 - 2(x)(22) + 22^2 = (x-22)^2$. So, the integral becomes: $I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}((x - 22)^2)}}dx} \quad \text{(Equation 1)}$ We can use the logarithm property $\log_e a^n = n \log_e a$ to simplify further: $I = \int\limits_6^{16} {{{2{{\log }_e}{x}} \over {{2{{\log }_e}{x} + 2{{\log }_e}(x - 22)}}}dx}$ $I = \int\limits_6^{16} {{{{{\log }_e}{x}} \over {{{\log }_e}{x} + {{\log }_e}(x - 22)}}}dx$ Using the property $\log_e a + \log_e b = \log_e (ab)$, the denominator can be written as $\log_e (x(x-22))$. $I = \int\limits_6^{16} {{{{{\log }_e}{x}} \over {{{\log }_e}(x(x - 22))}}}dx$ Let's re-examine the original form in terms of $x^2$ and $(x-22)^2$ to ensure we don't lose any information and to correctly apply King's Rule. $I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}((x - 22)^2)}}dx} \quad \text{(Equation 1 revisited)}$ We must ensure the arguments of the logarithms are positive. For $x \in [6, 16]$, $x^2 > 0$. Also, for $x \in [6, 16]$, $x-22$ is negative (ranging from $6-22=-16$ to $16-22=-6$), so $(x-22)^2$ is positive. Thus, the logarithms are well-defined.

Step 2: Apply King's Rule

We use the property $\int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx$. Here, $a=6$ and $b=16$, so $a+b = 6+16 = 22$. We replace $x$ with $(22-x)$ in the integrand: The new integrand will be: $\frac{{{{\log }_e}{(22-x)^2}}}{{\log_e{(22-x)^2} + \log_e{((22-x) - 22)^2}}}$ Let's simplify the terms:

• Numerator: $\log_e (22-x)^2$
• Denominator term 1: $\log_e (22-x)^2$
• Denominator term 2: $\log_e ((22-x) - 22)^2 = \log_e (-x)^2 = \log_e x^2$

So, applying King's Rule, the integral $I$ can also be written as: $I = \int\limits_6^{16} {{{{{\log }_e}{(22-x)^2}} \over {{{\log }_e}{(22-x)^2} + {{\log }_e}{x^2}}}}dx \quad \text{(Equation 2)}$

Step 3: Combine the Integrals

Now, we add Equation 1 and Equation 2: $I + I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}((x - 22)^2)}}dx} + \int\limits_6^{16} {{{{{\log }_e}{(22-x)^2}} \over {{{\log }_e}{(22-x)^2} + {{\log }_e}{x^2}}}}dx$ 2I = \int\limits_6^{16} {\frac{{{{\log }_e}{x^2}} + {{{\log }_e}{(22-x)^2}}}}{{{{\log }_e}{x^2} + {{\log }_e}{(22-x)^2}}}}dx The numerator and denominator of the integrand are identical. Therefore, the integrand simplifies to 1. $2I = \int\limits_6^{16} 1 \,dx$

Step 4: Evaluate the Simplified Integral

Now, we evaluate the integral of 1 with respect to $x$: $2I = [x]_6^{16}$ $2I = 16 - 6$ $2I = 10$

Step 5: Solve for I

Finally, we solve for $I$: $I = \frac{10}{2}$ $I = 5$

Common Mistakes & Tips

• Incorrectly applying logarithm properties: Ensure that the arguments of the logarithms are positive. In this problem, squaring the terms like $(x-22)^2$ ensures positivity, but careful attention to the domain of the original terms is crucial.
• Algebraic errors in simplifying $(a+b-x)$: Double-check the substitution of $(a+b-x)$ into the integrand, especially within complex expressions like logarithms.
• Forgetting to divide by 2: After adding the two forms of the integral, remember that you have $2I$, and the final answer for $I$ requires dividing by 2.

Summary

The problem involves a definite integral with a logarithmic function in the integrand. The key to solving this problem efficiently is to recognize the structure of the denominator and apply the King's Rule property of definite integrals. By setting $a=6$ and $b=16$, we found that $a+b=22$. Applying King's Rule transforms the integral into a form where, when added to the original integral, the numerator and denominator of the integrand become identical, simplifying to 1. Integrating 1 over the given limits yields the value of $2I$, from which we can easily find $I$.

The final answer is \boxed{5}.