Question

\(\int_2^4 \frac{\log x^2}{\log x^2+\log \left(36-12 x+x^2\right)} \mathrm{d} x\) is equal to

• A. 1
• B. 2
• C. 4
• D. 6

Hint:

Use \(\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\) and add the two forms.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Since \(36-12x+x^2=(6-x)^2\), the integrand is \(\dfrac{\log x^2}{\log x^2+\log(6-x)^2}\).

Step 2: Apply the Property
Here \(a+b=6\), so replacing x by \(6-x\) gives \(I=\displaystyle\int_2^4\dfrac{\log(6-x)^2}{\log(6-x)^2+\log x^2}dx\).

Step 3: Add the Two Forms
\(2I=\displaystyle\int_2^4 1\,dx=2\)

Step 4: Result
\(I=1\).

Final Answer:
The value of the integral is 1. \[ \boxed{\text{(A) }1} \]