यदि f_n(x) = log log log ldots log x (जहाँ lo | vedclass

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(A) दिया गया है $f_n(x) = \log \log \ldots \log x$ ($n$ बार)। माना $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$। माना $t = f_n(x) = \log(f_{n-1

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

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(A) दिया गया है $f_n(x) = \log \log \ldots \log x$ ($n$ बार)। माना $I = \int \frac{dx}{x f_1(x) f_2(x) \ldots f_n(x)}$। माना $t = f_n(x) = \log(f_{n-1}(x))$। तब, $\frac{dt}{dx} = \frac{1}{f_{n-1}(x)} \cdot \frac{d}{dx}(f_{n-1}(x)) = \frac{1}{f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x}$। अतः, $dx = (x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt$। इस मान को समाकलन में प्रतिस्थापित करने पर: $I = \int \frac{(x f_1(x) f_2(x) \ldots f_{n-1}(x)) dt}{x f_1(x) f_2(x) \ldots f_{n-1}(x) f_n(x)} = \int \frac{dt}{f_n(x)} = \int \frac{dt}{t}$। $I = \log(t) + c = \log(f_n(x)) + c$। चूंकि $f_{n+1}(x) = \log(f_n(x))$, इसलिए $I = f_{n+1}(x) + c$।

यदि $f_n(x) = \log \log \log \ldots \log x$ (जहाँ $\log$ $n$ बार दोहराया गया है), तो $\int (x f_1(x) f_2(x) \ldots f_n(x))^{-1} dx$ का मान ज्ञात कीजिए।

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