How to solve this finite difference equation related to falling factorials? I am following Gleich(2005): "Finite Calculus: A Tutorial for Solving Nasty Sums". At one point when referring to the fact that

The function $H_x$ (https://en.wikipedia.org/wiki/Harmonic_number) is
  the antiderivative of $x^{\underline{-1}}$ (The underline is a notation to write a falling factorial, $x^{\underline{-1}} = \dfrac{1}{x+1}$).

The author writes

There is no easy way of getting the correct function ($H_x$) intuitively. 

So I am wondering what is the way of getting the correct function? I realized that basically one has to solve the following finite difference equation for $f(x)$:
$$
f(x+1) - f(x) = \frac{1}{x+1}.
$$
But, apart from guessing the correct function, is there a systematic way of solving such an equation to see that the solution is $f(x) = H_x$?
Update: This finite difference equation is of course the discrete version of
$$
\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \rightarrow 0} \frac{1}{x+h}
$$ 
or short $\frac{df}{dx} = 1/x$ with the solution $f(x) = \ln(x) + C$.
 A: By telescoping,$$
f(n) = f(0) + \sum_{k = 1}^n (f(k) - f(k - 1)) = f(0) + \sum_{k = 1}^n \frac{1}{k} = f(0) + H_n.
$$
A: Note that the solution is determined up to an arbitrary periodic summand function $P_1(x)$ with period $1.$
Then, applying Laplace transformation 
$$\mathcal L(f(x)) = F(s)$$
to the issue equation in the form of
$$f(x)-f(x-1) = \dfrac1x,$$
one can get
$$F(s) = \dfrac1{1-e^{-s}} = \sum\limits_{n=0}^\infty e^{-ns},$$
$$f(x) = \sum\limits_{n=0}^\infty \dfrac{u(x-n)}{x-n}+P_1(x),\quad x \ge 1,$$
$$f(x) = \sum\limits_{n=1}^\infty \dfrac{u(x-n)}{x-n}+P_1(x),$$
where
$$u(x) =\begin{cases}
0,\text{ if } x<0\\
1,\text{ if } x\ge0.\\
\end{cases}$$

Addition of 12.06.18
Let us consider the case of the integer positive $x$ with the initial condition
$$f(1) = 1.$$
Then
$$P_1(x) = C + Z_1(x),$$
where $C$ is the constant, which can be defined from the initial conditions, and
$Z_1(x)$ is the arbitrary periodic function, which satisfy the condition $Z_1(1)= 0.$
For example, $Z_1(x)$ can be choosen in the form of
$$Z_1(x) = \sum\limits_{i=1}^{10} A_i\sin2\pi ix,$$
where $A_i$ are the arbitrary coefficients.
So
$$f(1) = 1 + C + Z_1(1),$$
$$1 = 1 + C,$$
$$ C= 0,$$
$$f(x) = \sum\limits_{n=1}^\infty \dfrac{u(x-n)}{x-n} + Z_1(x),\quad x\ge1,$$
$$f(m) = \sum\limits_{k=1}^m \dfrac1k,\quad m\in\mathbb N.$$
