I was looking for the asymptotic behaviour of the anti-derivative of $\frac{1}{\ln \ln x}$, in terms of the big-O notation.
Wikipedia's list does not have this integral, and Wolfram Alpha says "no results found in terms of standard mathematical functions".
For the similar function $\frac{1}{\ln x}$, the anti-derivative is asymptotically $\Theta(\frac{x}{\ln x})$. So it makes sense to assume (as Will Jagy commented) that here, too, the antiderivative is $\Theta(\frac{x}{\ln \ln x})$. This raises a more general question:
In what cases is this true that:
$$ \int{f(x)dx} = \Theta(x f(x)) $$
?
That worked, integration by parts, repeat while patience lasts and the things being integrated are getting smaller; $$ u = \frac{1}{\log \log x}, \; \; \; dv = 1 \, dx $$ $$ du = \frac{-1}{x \log x \left( \log \log x \right)^2} dx, \; \; v = x,$$ $$ \int \frac{1}{\log \log x} \; dx = \frac{x}{\log \log x} + \int \frac{1}{ \left( \log \log x \right)^2 \; \log x} dx $$
thinking about next step, but we can already see that the remaining integral is smaller than the original.
EDDDITTT: it seems repeating with $dv = 1 dx$ is the way to go. $$ \int \frac{1}{\log \log x} \; dx = \frac{x}{\log \log x} + \frac{x}{ \left( \log \log x \right)^2 \; \log x} + \int \frac{1 + 2 \log \log x}{ 2 \; \left( \log \log x \right)^3 \; \left( \log x \right)^2} dx $$