I've to find (the average of a function over a particular interval, where $t_1>0$, $t_2>0$ and $t_2>t_1$):
$$\frac{1}{t_2-t_1}\int_{t_1}^{t_2}x(t)dt\tag1$$
Where $x(t)$ is the solution to the following DE (with intial condition $x(0)=x_0)$:
$$x(t)\cdot r+x'(t)\cdot l+a\cdot\ln\left(1+\frac{x(t)}{b}\right)=0\space\Longleftrightarrow\space x(t)=\dots\tag2$$
Now, according to the answer of @JJacquelin on my previuous question, I could write $x(t)$ as follows:
$$t=-l\int_{x_0}^{x(t)}\frac{d\xi}{r\xi+a\ln\left(1+\frac{\xi}{b}\right)}\tag3$$
But I do not see how that can help me find $(1)$?!
The following provides an implicit solution to you problem (this might be the most "explicit" solution to your problem given that you only know the implicit solution of the ODE):
Let us assume that $x(t)$ is monotonous on the interval $[t_1,t_2]$ (if not you have to apply the following reasoning on each interval on which $x(t)$ is monotonous. Let us change the variables from $x(t)$ to $\xi$. We obtain $$\frac{1}{t_2-t_1}\int_{t_1}^{t_2} x(t)\,dt = \frac{1}{t_2-t_1} \int_{t_1}^{t_2} \frac{\xi}{x'}\,dt =\frac{-l}{t_2-t_1}\int_{x(t_1)}^{x(t_2)}\frac{\xi\,d\xi}{r\xi+a\ln\left(1+\frac{\xi}{b}\right)}\,.$$