If l^r(x) denotes the r-th iterated logarithm | vedclass

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(A) Let $I = \int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx$. Consider the substitution $t = l^{r+1}(x)$. By the chain rule,t

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$l^{r+1}(x) + c$

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(A) Let $I = \int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx$. Consider the substitution $t = l^{r+1}(x)$. By the chain rule,the derivative of $l^{r+1}(x) = \log(l^r(x))$ is $\frac{d}{dx} l^{r+1}(x) = \frac{1}{l^r(x)} \cdot \frac{d}{dx} l^r(x)$. Continuing this process,we find that $\frac{d}{dx} l^{r+1}(x) = \frac{1}{l^r(x) \cdot l^{r-1}(x) \cdot ... \cdot l^1(x) \cdot x}$. Thus,$dt = \frac{1}{x \cdot l^1(x) \cdot ... \cdot l^r(x)} \, dx$. Substituting this into the integral,we get $\int 1 \, dt = t + c$. Therefore,the integral is $l^{r+1}(x) + c$.

If $l^r(x)$ denotes the $r$-th iterated logarithm of $x$,i.e.,$l^1(x) = \log(x)$,$l^2(x) = \log(\log(x))$,...,$l^r(x) = \log(\log(...\log(x)...))$,then $\int \frac{1}{x \cdot l^1(x) \cdot l^2(x) \cdot ... \cdot l^r(x)} \, dx = $

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