Let us calculate the first few iterated integrals of $$f(t) = \frac 1 t$$
It is well known from high school
$$I_1\{f,t\}(t) = \int \frac 1 t dt = \log(t) + C_1$$ Then in our first calculus course at University we maybe derive: $$I_2\{f,t\}(t) = \int \log(t) dt = t(\log(t)-1) + C_1t + C_2$$
We can furthermore verify :
$$I_3\{f,t\}(t) = \int t(\log(t)-1) dt = \frac{t^2}{4}(2\log(t)-3)+C_1t^2+C_2t+C_3$$
This seems to follow a pattern. Let $P_1,P_2,P_3$ be polynomials. We can now find them so that $P_1(t)P_2(\log(t)) + P_3(t)$ equals any of the above expressions and $P_3$ is created by integration constants.
Can we prove that any $N$ times iterated integral of $f(t) = \frac 1 t$ will follow this pattern?
$$I_0\{f,t\}(t_0) = f(t_0)\\I_N\{f,\xi\}(t_0) = \left(\int I_{N-1}\{f,\xi\} d\xi\right)(t_0)$$
Because $$\forall n\in\mathbb N,\int_0^t u^n\log(u)=\left[\dfrac{u^{n+1}}{n+1}\times \log(u)\right]_0^t-\int_0^t \dfrac{u^{n+1}}{n+1}\times \dfrac{1}{u}\text{d}u=\dfrac{t^{n+1}}{n+1}\log(t)-\dfrac{t^{n+1}}{(n+1)^2} $$
So if $f \in \mathbb R[x] \times \log(x)+\mathbb R[x]=L$ then $$g(x)=\int_0^x f(u)\text{d}u,g \in L$$