If f_n(x) = log log log ldots log x (where lo | vedclass

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(A) Given $f_n(x) = \log(\log(\ldots \log x))$ ($n$ times). Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$. Let $t = f_n(x) = \log(f_{n-1}(x)

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$f_{n+1}(x) + c$

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(A) Given $f_n(x) = \log(\log(\ldots \log x))$ ($n$ times). Let $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$. Let $t = f_n(x) = \log(f_{n-1}(x))$. Then,$\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$. Thus,$dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$. Substituting this into the integral: $I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)}$. Wait,the substitution $t = f_n(x)$ implies $dt = \frac{dx}{x f_1(x) \ldots f_{n-1}(x)}$. Therefore,$I = \int \frac{dt}{t} = \log|t| + c = \log|f_n(x)| + c$. Since $f_{n+1}(x) = \log(f_n(x))$,we have $I = f_{n+1}(x) + c$.

If $f_n(x) = \log \log \log \ldots \log x$ (where $\log$ is repeated $n$ times),then $\int (x f_1(x) f_2(x) \ldots f_n(x))^{-1} dx$ is equal to

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