Consider the function $$f_\epsilon(x):=\int_0^1 \frac{(1-z)^x-1}{\log(1-\epsilon z)}\, dz,$$ defined for a parameter $\epsilon \in (0,1]$ and $x \geq 0$. When $\epsilon=1$, this is $\log(1+x)$. One way to see this is to take the derivative of the function with respect to $x$ and see that the resulting integral evaluates to $1/(1+x)$, and moreover notice that $f_1(0)=0$.

In general, one can express the function as $$f_\epsilon(x)=\frac{1}{\epsilon}\log(1+x)+E_\epsilon(x).$$ It can be observed that, as $x$ tends to infinity, $E_\epsilon(x)$ converges to a constant depending on $\epsilon$. A plot of $C(\epsilon):=\lim_{x \to \infty} E_\epsilon(x)$, as a function of $\epsilon$ is given below: And here is a plot of $\epsilon C(\epsilon)$, which seems to converge to the Euler–Mascheroni constant as $\epsilon\to 0$: What is the value of $C(\epsilon)$? Can it be expressed in terms of common special functions (such as logarithmic integral)?