This is not particularly a useful integral or one asked in an exam or anything. I just really enjoy doing random integrals and derivatives. With that being said, how can I find:
$$\int \left(1+\ln \left(1+\ln (...+\left(1+ \ln(x))\right)\right)\right) dx$$
I tried to star with the simple case of: $$\int \left(1+\ln(x)\right)dx=x\ln \left(x\right)+C$$ I then introduced the first nest. Here I did Integration by Parts with $u=1+\ln \left(1+\ln \left(x\right)\right), v'=1$: $$\int \left(1+\ln \left(1+\ln \left(x\right)\right)\right)dx=x\left(1+\ln \left(1+\ln \left(x\right)\right)\right)-\frac{1}{e}\text{Ei}\left(\ln \left(x\right)+1\right)+C$$
However, these take quite some time and it's only the first nest. Therefore, I'm curious as to if $(a)$ such an integral does have solution and $(b)$ how to derive that situation.
PS - Again this is just a fun question and not from an exam, website or anywhere important.
$$\int 1+\ln(1+\ln(1+\ln(1+...\ln(x)))) dx=F(x)$$
$$1+\ln(1+\ln(1+\ln(1+...\ln(x))))=F'(x)$$
$$1+\ln(F'(x))=F'(x)$$
$$F'(x)=1$$
$$F(x)=x+c$$