Question

The value of the integral \(\int_{5}^{9} \frac{\log(3x^2)}{\log(3x^2) + \log(588 - 84x + 3x^2)} \, dx\) is equal to:

• A. 2
• B. 1
• C. $\frac{1}{2}$
• D. 4

Hint:

For integrals of the form $\int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} \, dx$, the result is always $\frac{b-a}{2}$. In this case: $\frac{9-5}{2} = \frac{4}{2} = 2$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem utilizes the Generalized King’s Property for definite integrals over an interval \([a, b]\).
The integrand has a specific symmetric structure: \(\frac{f(x)}{f(x) + f(a+b-x)}\).
When we see such logarithmic expressions with quadratic terms in the denominator, we should immediately check if they represent the reflection of the numerator function.
Step 2: Key Formula or Approach:
Rule: \(\int_{a}^{b} \frac{f(x)}{f(x) + f(a+b-x)} dx = \frac{b-a}{2}\).
We must verify if the second term in the denominator is indeed \(f(a+b-x)\).
Step 3: Detailed Explanation:
Let \(f(x) = \log 3x^{2}\). The limits are \(a = 5\) and \(b = 9\).
Thus, \(a + b = 14\). We need to calculate \(f(14 - x)\):
\[ f(14 - x) = \log [3(14 - x)^{2}] = \log [3(196 - 28x + x^{2})] \]
\[ f(14 - x) = \log [588 - 84x + 3x^{2}] \]
This matches the second term in the denominator. Let the integral be \(I\).
Applying the property \(\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx\):
\[ I = \int_{5}^{9} \frac{\log(588 - 84x + 3x^{2})}{\log(588 - 84x + 3x^{2}) + \log 3x^{2}} dx \]
Adding the original \(I\) and the modified \(I\):
\[ 2I = \int_{5}^{9} \frac{\log 3x^{2} + \log(588 - 84x + 3x^{2})}{\log 3x^{2} + \log(588 - 84x + 3x^{2})} dx \]
\[ 2I = \int_{5}^{9} 1 dx \implies 2I = [x]_{5}^{9} = 9 - 5 = 4 \]
\[ I = \frac{4}{2} = 2 \]
Step 4: Final Answer:
The value of the integral is 2.