$f_n(x)=\log \cdot \log \cdot \log \ldots \cdot \log x$ (upto $n$ times)
$f_1(x)=\log x$
$f_2(x)=\log \log x$
$f_3(x)=\log \log \log x$
$f_{n-1}(x)=\log \log \log \ldots \ldots \log x$
(up to $(n-1)$ times)
Now, $\int\left(x f_1(x) f_2(x) \ldots f_n(x)\right)^{-1} d x$
$=\int \frac{d x}{\left[x f_1(x) f_2(x) \ldots f_n(x)\right]}$
$=\int \frac{\left[x f_1(x) f_2(x) \ldots f_{n-1}(x)\right] d t}{\left[x f_1(x) f_2(x) \ldots f_{n-1}(x)\right] \cdot t}=\int \frac{d t}{t}$
$=\log t+c$
$\left[\begin{array}{l}\text { Put, } f_n(x)=t \\ \frac{d t}{d x}=\frac{1}{\left[f_{n-1}(x) f_{n-2}(x) \ldots f_1(x) \cdot x\right]} \\ d x=\left[x f_1(x) \cdot f_2(x) \ldots f_{n-1}(x)\right] d t\end{array}\right]$
$=\log f_n(x)+c$
$=f_{n+1}(x)+c$
Hence,
$\int\left(x f_1(x) f_2(x) \ldots f_n(x)\right)^{-1} d x=f_{n+1}(x)+c$