Key Concepts and Formulas

• King Property of Definite Integrals: For a definite integral $\int_a^b f(x) \,dx$, the property states that $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$. This is useful when the integrand $f(x)$ exhibits symmetry or a complementary relationship with $f(a+b-x)$.
• Logarithm Properties: $\log_e e^k = k$.
• Substitution Rule for Definite Integrals: If $u = g(x)$, then $du = g'(x) \,dx$. The limits of integration change from $a$ and $b$ to $g(a)$ and $g(b)$ respectively.

Step-by-Step Solution

Let the given integral be $I$. $I = \int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right) d x$

Step 1: Identify the limits of integration and the integrand. The limits of integration are $a = e^2$ and $b = e^4$. The integrand is $f(x) = \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _e x\right)^2+1\right)^{-1}}}\right)$.

Step 2: Apply the King Property of Definite Integrals. The King Property states $\int_a^b f(x) \,dx = \int_a^b f(a+b-x) \,dx$. In this case, $a+b = e^2 + e^4$. Let's consider a substitution $y = a+b-x$. Then $dy = -dx$. However, it is more direct to substitute $x$ with $a+b-x$ in the integrand. The property is most useful when we add the original integral to the transformed integral. Let's analyze the term $a+b-x$ in the context of the logarithm. If we replace $x$ with $a+b-x$, the term $\log_e x$ becomes $\log_e (a+b-x)$. This doesn't seem to simplify nicely.

Let's re-examine the structure of the integrand. It involves $\log_e x$ and $6 - \log_e x$. This suggests a different approach might be more suitable or that the King Property needs to be applied with a specific transformation.

Consider the transformation $u = \log_e x$. Then, $du = \frac{1}{x} \,dx$. When $x = e^2$, $u = \log_e e^2 = 2$. When $x = e^4$, $u = \log_e e^4 = 4$. The integral becomes: $I = \int_2^4 \left(\frac{e^{\left(u^2+1\right)^{-1}}}{e^{\left(u^2+1\right)^{-1}}+e^{\left(\left(6-u\right)^2+1\right)^{-1}}}\right) du$ Let $g(u) = \frac{e^{\left(u^2+1\right)^{-1}}}{e^{\left(u^2+1\right)^{-1}}+e^{\left(\left(6-u\right)^2+1\right)^{-1}}}$. The integral is $I = \int_2^4 g(u) \,du$.

Step 3: Apply the King Property to the transformed integral. Now, we apply the King Property $\int_a^b g(u) \,du = \int_a^b g(a+b-u) \,du$ to this integral. Here, $a=2$ and $b=4$. So, $a+b = 2+4 = 6$. We need to evaluate $g(6-u)$: $g(6-u) = \frac{e^{\left((6-u)^2+1\right)^{-1}}}{e^{\left((6-u)^2+1\right)^{-1}}+e^{\left(\left(6-(6-u)\right)^2+1\right)^{-1}}}$ $g(6-u) = \frac{e^{\left((6-u)^2+1\right)^{-1}}}{e^{\left((6-u)^2+1\right)^{-1}}+e^{\left(u^2+1\right)^{-1}}}$ Let's add the original integral $I$ and the integral with $g(6-u)$: $I = \int_2^4 g(u) \,du$ $I = \int_2^4 g(6-u) \,du$ Adding these two expressions for $I$: $2I = \int_2^4 g(u) \,du + \int_2^4 g(6-u) \,du$ $2I = \int_2^4 [g(u) + g(6-u)] \,du$

Step 4: Simplify the sum of the integrands. Let's compute $g(u) + g(6-u)$: $g(u) + g(6-u) = \frac{e^{\left(u^2+1\right)^{-1}}}{e^{\left(u^2+1\right)^{-1}}+e^{\left(\left(6-u\right)^2+1\right)^{-1}}} + \frac{e^{\left((6-u)^2+1\right)^{-1}}}{e^{\left((6-u)^2+1\right)^{-1}}+e^{\left(u^2+1\right)^{-1}}}$ The denominators are the same. Let $A = e^{\left(u^2+1\right)^{-1}}$ and $B = e^{\left((6-u)^2+1\right)^{-1}}$. Then $g(u) = \frac{A}{A+B}$ and $g(6-u) = \frac{B}{B+A}$. So, $g(u) + g(6-u) = \frac{A}{A+B} + \frac{B}{A+B} = \frac{A+B}{A+B} = 1$.

Step 5: Evaluate the integral of the simplified sum. Now, the integral becomes: $2I = \int_2^4 1 \,du$ $2I = [u]_2^4$ $2I = 4 - 2$ $2I = 2$

Step 6: Solve for I. $I = \frac{2}{2}$ $I = 1$

Common Mistakes & Tips

• Incorrect Application of King Property: Ensure the limits of integration $a$ and $b$ are correctly identified and used in the expression $a+b-x$ or $a+b-u$.
• Mistakes in Substitution: When using substitution, remember to change the limits of integration according to the substitution made.
• Algebraic Errors: Be careful with exponents and algebraic manipulations, especially when dealing with terms like $(u^2+1)^{-1}$ and $(6-u)^2$.

Summary

The problem is solved by first performing a substitution $u = \log_e x$ to simplify the integrand and change the limits of integration. The transformed integral is then subjected to the King Property of Definite Integrals, $\int_a^b g(u) \,du = \int_a^b g(a+b-u) \,du$. By adding the original integral and the transformed integral, the sum of the integrands simplifies to a constant, making the integration straightforward. The final result is obtained by solving for the original integral $I$.

The final answer is $\boxed{1}$. This corresponds to option (A).