Unfortunately you cannot expect to have anything sufficiently simple. Suppose you have
$$\int f(x) dx = F(x)$$
Now we would like to find the connection $T$
$$\int \ln(f(x)) dx = T(F(x))$$
But look:
$$\ln(f(x)) = T'(F(x))f(x)$$
However, $T'(F(x)) \cdot f(x)$ is multiplicative to $f(x)$ and you have $\ln(f(x))$ to the left. We know no such expression that would be simple to state. On top of that you need some operator over $F(x)$ which is an integral(!). That simply will not do it.
To the rescue, you can simply reverse integration and keep on as far as you can:
$$\int \ln(f(x)) dx = x\ln(f(x)) + g(x)$$
$$\ln(f(x)) = \frac{x}{f(x)}f'(x) + \ln(f(x)) + g'(x)$$
This is making:
$$g'(x) = -\frac{x}{f(x)}f'(x)$$
$$g(x) = -\int x \, d\ln(f(x))$$
Now comes the funny part. I will show it for a triple and we can easily derive the formula:
$$d\ln\ln\ln(x)=\frac1{x\ln(x)\ln(\ln(x))}$$
It is simply
$$d\ln^{(n)}(x)=\frac1{x\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)}$$
or
$$\int \ln^{(n)}(x) dx = x\ln^{(n)}(x) - \int\frac1{\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)} \, dx$$
I know it sounds disappointing, and nobody can say if some theory can actually do something about this, but I very much doubt that anything better can be done in general.
However, notice, that in many of these cases you can use the telescoping rule, for this situation suitably adjusted to:
$$ \left ( x\sum_{n=0}^{+\infty}{f_n(x)} \right )' = x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=0}^{+\infty}{f_n(x)} $$
Now take
$$f_{n+1}(x) = - xf_n'(x), f_0(x)=\ln^{(k)}(x)$$
and you would telescopically cancel it all:
$$x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=0}^{+\infty}{f_n(x)} = x\sum_{n=0}^{+\infty}{f_n'(x)} + \sum_{n=1}^{+\infty}{f_n(x)} + f_0(x) = $$ $$x\sum_{n=0}^{+\infty}{f_n'(x)} - x\sum_{n=0}^{+\infty}{f_n'(x)} + f_0(x) = \ln^{(k)}(x)$$
So it makes
$$ \left ( x\sum_{n=0}^{+\infty}{f_n(x)} \right )' = \ln^{(k)}(x) $$
or
$$ \int \ln^{(k)}(x) \, dx = x \left (\ln^{(k)}(x) - \frac1{\prod\limits_{k=1}^{n-1}\ln^{(k)}(x)} + \sum_{n=2}^{+\infty}{f_n(x)} \right )$$
repeating the recursion
$$f_0(x)=\ln^{(k)}(x), f_{n+1}(x) = - xf_n'(x)$$
By this, at least you got rid of the integral sign. Important here is that the recursion is giving all slower and slower function, which is the case here.
This is adjusted from a more general approach (of the same author, so it is still all attempted contribution of mhe [my hamble existence]):
https://math.stackexchange.com/a/4186998/944732
In general you can find that
$$\int \ln^{(k)}(x) \, dx = x \left (\ln^{(k)}(x) + \sum_{n=1}^{k-1}{\frac{(-1)^n}{\ln^{n}(x)}}\sum_{n \geq m_1 \geq m_2... \geq m_{k-1} \geq 1}{\frac{c_{n,m_1,...,m_{k-1}}}{\prod\limits_{j=1}^{k-1}(\ln^{(j+1)}(x))^{m_j}}} \right )$$
where constants $c_{n,m_1,...,m_{k-1}}$ are recursively defined, and combinatorial in nature:
$$c_{k_1,k_2,k_3,...,k_m}=\sum_{n=1}^{m}c_{k_1-\theta(1,n),k_2-\theta(2,n),k_3-\theta(3,n),...,k_m-\theta(m,n)} (k_n-1)$$
$$\theta(i,j)=0 \text{, except } \theta(i,j)=1 \text{ if } i \leq j$$
where additionally
$$c_{k_1,k_2,k_3,...,k_m}=0 \text{ if any } k_i < k_j \text{ when } i<j \text{ or } k_i \leq 0$$
and
$$c_{1,...,1}=1$$
