Let \(f(x) = x\), \(f_1(x) = f(logx)\), \(f_2(x) = f_1(logx)\), \(f_3(x) = f_2(logx)\), \(\ldots\) and so on. Then \(\int \frac{1}{f(x)\,f_1(x)\,f_2(x)\,\ldots f_{2026}(x)}\,dx = \ldots\)
Show Hint
Differentiate f_k(x) = log log ... log x and look for the pattern 1/(f f1 ... f(k-1)).
Step 1: Try a small case:
Take only $f\cdot f_1 = x\log x$. Then $\int\frac{dx}{x\log x} = \log\log x = f_2$. The product ends at $f_1$ and the answer is $f_2$, one index higher.
Step 2: Check the next case:
Take $f f_1 f_2 = x\log x\log\log x$. Substituting $u = \log\log x$ gives $\int\frac{du}{u} = \log u = \log\log\log x = f_3$. Again one index higher.
Step 3: Generalise:
The product ends at $f_{n}$ and the integral is $f_{n+1}$. Here $n = 2026$, so the integral is $f_{2027}(x) + c$.
Final Answer:
Option (C).
\[ \boxed{f_{2027}(x)+c \text{ (C)}} \]