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

Concept: To solve the problem, we use King's Property of definite integrals: $$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$$ This property is particularly useful when the denominator remains unchanged or symmetrically transforms under the substitution $x \rightarrow (a+b-x)$.

Step 1: Analyze the integrand.
Let $I = \int_{5}^{9} \frac{\log 3x^2}{\log 3x^2 + \log(588 - 84x + 3x^2)} \, dx$.
Notice the term $588 - 84x + 3x^2$. We can factor out 3: $3(196 - 28x + x^2) = 3(14 - x)^2$.
So the integral is $I = \int_{5}^{9} \frac{\log 3x^2}{\log 3x^2 + \log 3(14 - x)^2} \, dx$.

Step 2: Apply King's Property.
Substitute $x$ with $(5 + 9 - x) = (14 - x)$: $$I = \int_{5}^{9} \frac{\log 3(14-x)^2}{\log 3(14-x)^2 + \log 3(14 - (14-x))^2} \, dx$$ $$I = \int_{5}^{9} \frac{\log 3(14-x)^2}{\log 3(14-x)^2 + \log 3x^2} \, dx$$

Step 3: Sum the integrals.
Adding the two forms of $I$: $$2I = \int_{5}^{9} \frac{\log 3x^2 + \log 3(14-x)^2}{\log 3x^2 + \log 3(14-x)^2} \, dx$$ $$2I = \int_{5}^{9} 1 \, dx$$ $$2I = [x]_{5}^{9} = 9 - 5 = 4$$ $$I = \frac{4}{2} = 2$$