Question

$\int_{6}^{16} \frac{\log_e x^2}{\log_e x^2 + \log_e(x^2 - 44x + 484)} \, dx$ is equal to :

• A. $10$
• B. $8$
• C. $6$
• D. $5$

Hint:

If the denominator is of the form $f(x) + f(a+b-x)$, the value of the integral is always $(b-a)/2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Use the King's Property of definite integrals: $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$.
Step 2: Detailed Explanation:
Let $I = \int_{6}^{16} \frac{\ln x^2}{\ln x^2 + \ln(x^2 - 44x + 484)} \, dx$.
The term in the denominator is $x^2 - 44x + 484 = (x - 22)^2$.
So $I = \int_{6}^{16} \frac{\ln x^2}{\ln x^2 + \ln(x - 22)^2} \, dx$.
Using the property $\int_a^b f(x) \, dx$ with $a=6, b=16$, so $a+b = 22$:
Replace $x$ with $22 - x$:
\[ I = \int_{6}^{16} \frac{\ln (22-x)^2}{\ln (22-x)^2 + \ln(22-x-22)^2} \, dx = \int_{6}^{16} \frac{\ln(x-22)^2}{\ln(x-22)^2 + \ln x^2} \, dx \]
Adding the two forms of $I$:
\[ 2I = \int_{6}^{16} \frac{\ln x^2 + \ln(x-22)^2}{\ln x^2 + \ln(x-22)^2} \, dx \]
\[ 2I = \int_{6}^{16} 1 \, dx = [x]_6^{16} = 16 - 6 = 10 \]
\[ I = 5 \]
Step 3: Final Answer:
The integral evaluates to $5$.