Question:
The integral
\( \displaystyle \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log (36 - 12x + x^2)}\,dx \)
is equal to :
The integral
\( \displaystyle \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log (36 - 12x + x^2)}\,dx \)
is equal to :
\( \displaystyle \int_{2}^{4} \frac{\log x^2}{\log x^2 + \log (36 - 12x + x^2)}\,dx \)
is equal to :
Show Hint
Simplify \(36-12x+x^2=(6-x)^2\) and apply \(\int_a^b f(x)dx=\int_a^b f(a+b-x)dx\) with \(a+b=6\).
Updated On: Nov 11, 2026
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The Correct Option is A
Solution and Explanation
For \(2\le x\le 4\): \(\log x^2 = 2\log x\) and \(\log(36-12x+x^2) = \log (6-x)^2 = 2\log(6-x)\).
So the integrand is \(f(x)=\frac{\log x}{\log x+\log(6-x)}\) (the denominator is \(\log(x(6-x))\gt 0\) since \(x(6-x)\ge 8\)).
Let \(I=\int_2^4 f(x)dx\). Using \(x\to 6-x\) (since \(2+4=6\)): \(I=\int_2^4 \frac{\log(6-x)}{\log(6-x)+\log x}dx\).
Adding: \(2I = \int_2^4 1\,dx = 2 \Rightarrow I = 1\).
Answer: (1) 1
So the integrand is \(f(x)=\frac{\log x}{\log x+\log(6-x)}\) (the denominator is \(\log(x(6-x))\gt 0\) since \(x(6-x)\ge 8\)).
Let \(I=\int_2^4 f(x)dx\). Using \(x\to 6-x\) (since \(2+4=6\)): \(I=\int_2^4 \frac{\log(6-x)}{\log(6-x)+\log x}dx\).
Adding: \(2I = \int_2^4 1\,dx = 2 \Rightarrow I = 1\).
Answer: (1) 1
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