General Solution of Functional Equation What is the general solution to:
$$
\frac{f(x+h)-f(x)}{h} = \frac{1}{x} \tag{1}
$$
Obviously the solution to this for the limiting case of $h\to 0$ is 
$f(x) = \ln(x) + c$
Attempting to solve the case of $h=1$ poses some difficulties.
But I imagine that once I can cover the case of $h=1$ the rest shall become simpler
A related problem is that the solution to:
$$
\frac{f(x+h)-f(x)}{h} = f(x) \tag{2}
$$
Can be assumed to have exponential form. Yielding:
$$a^{x+h} - a^{x} = ha^x \rightarrow a^h = 1+h \rightarrow a=(1+h)^{\frac{1}{h}}$$
Thus the general solution to this similar problem is:
$$f = (1+h)^{\frac{x}{h}}$$

Obviously the general solution for $f(x+1)-f(x)=1/x$ (assuming $f(x)$ is required to be defined for $x>0$) is as follows. $f(x)=f_0(x)$ can be an arbitrary function for $x\in(0,1]$ and then, for noninteger $x$, 
$$ 
 f(x) =f_0(x-[x])+\sum_{n=1}^{[x]}\frac 1{x-n},
$$
where $[x]$ is the integer part of $x$, while 
$$
f(n)=f_0(1)+\sum_{k=1}^{n-1}\frac1k,\qquad n=2,3,\dots\,.
$$
In particular, the "general solution" in the OP is not general but only special. 
