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dxx(logx)(loglogx)(loglog8 times x) is equal to

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a
(log⁡log⁡…⏟8 times x)+C
b
(log⁡log⁡…⏟7 times x)+C
c
(log⁡log⁡…⏟9 times x)+C
d
None of these

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detailed solution

Correct option is C

Let   I=∫dxx(log⁡x)(log⁡log⁡x)…(log⁡log⁡…⏟8 times x)put , (log⁡log⁡…⏟8 times x)=t⇒1(log⁡log⁡…⏟7 times x)…(log⁡log⁡x)(log⁡x)xdx=dt∴ I=∫dtt=log⁡t+C=(log⁡log⁡log⁡…⏟9 times x)+C


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