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04/01/2022 1:29 pm
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The value of \(\int_6^{16}\frac{ln \ x^2}{ ln \ x^2 + ln(x^2-44x + 484)}dx\) is
(a) 5
(b) 8
(c) 6
(d) 10
1 Answer
1
04/01/2022 1:44 pm
Correct answer: (a) 5
Explanation:
I = \(\int_6^{16}\frac{ln \ x^2}{ ln \ x^2 + ln(x - 22)^2}dx\)
= \(\int_6^{16}\frac{ln (22 -x)^2}{ ln (22 - x)^2 + ln x^2}dx\)
= \(\int_6^{16}\frac{ln (x -22)^2}{ ln x^2 + ln(22 - x)^2} dx\) (∴ \(\int_a^b f(x) \) = \(\int_a^b f(a + b - x)\))
= \(\int_6^{16}\frac{ln (x - 22)^2}{ ln x^2+ ln (x - 22)^2}dx\)
Thus I = \(\int_6^{16}\frac{ln \ x^2}{ ln \ x^2 + ln(x^2-44x + 484)}dx\)
= \(\int_6^{16}\frac{ln (22 -x)^2}{ ln (22 - x)^2 + ln x^2}dx\)
⇒ 2I = \(\int_6^{16}\frac{ln\ x^2 + ln (x -22)^2}{ ln x^2 + ln (x - 22)^2}dx\)
⇒ 2I = 16 - 6
⇒ I = 5