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JEE Mains · Maths · STD 12 - 7.2 definite integral

\(\int \limits_{6}^{16} \frac{\log _{\mathrm{e}} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x\) is equal to:

  1. A \(6\)
  2. B \(8\)
  3. C \(5\)
  4. D \(10\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

Let \(I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}\left(x^{2}-44 x+484\right)} d x\) \(I=\int_{6}^{16} \frac{\log _{e} x^{2}}{\log _{e} x^{2}+\log _{e}(x-22)^{2}} d x \ldots(1)\) We know \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x)\, d x(\text { king })\) So…
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