I recently completed a quiz for my Calc 2 class and I could not find the answer to this integral:
$$ \int \frac{1}{\left(\ln x\right)^{\ln x}}dx $$
I tried to use integration by parts and was able to get to: $$ \int \frac{1}{\left(\ln x\right)^{\ln x}}dx\:=\:\frac{n}{\left(\ln x\right)^{\ln x}}\:+\int \:\frac{\ln\left(\ln x\right)-1}{\left(\ln x\right)^{\ln x}}dx $$ but I don't know where to go from here.
You wrote "I don't know where to go from here". In fact, you are going to nowhere since your problem is still more difficult than the impossible Somophore dream.
If you let $x=e^t$, you face the problem of $$I=\int t^{-t} \,e^t\,dt$$
Even if you expand $$ t^{-t}=\sum_{n=0}^\infty (-1)^n \frac{ [t \, \log(t)]^n} {n!}$$ you should face integrals $$J_n=\int e^t [t \, \log(t)]^n\,dt$$
Just have a look at the result given by Wolfram Alpha for $J_2$.
Now, what you could do is to expand the whole integrand $$ t^{-t}\,e^t=\sum_{n=0}^\infty (-1)^n\frac {\big[\log(t)-1\big]^n} {n!} \,t^n$$ and $$K_n=\int \big[\log(t)-1\big]^n\,t^ n\,dt$$ can easily be integrated.
A small example $$\int_0^e t^{-t} \,e^t\,dt=\sum_{n=1}^\infty \frac {e^n}{n^n} $$ converges quite fast.
\ln xinstead oflnxwill appear as $\ln x$ instead of $lnx$. – Barry Cipra Jul 16 '21 at 02:58