How to solve Stieltjes integrals? I am confused on how to solve Stieltjes integrals because the $dx$ part in the integrals normally becomes $dg(x)$. 
For example, I have the solution to this but I just have a question about it:
let $g(x)=[x]$, the greatest integer $n$ such that $n \le x$. Evaluate the Stieltjes integral 
$$\int_0^3 g(x)d\sqrt{1+x^2}$$
after a few steps I end up at 
$$\int_1^2 d\sqrt{1+x^2} + 2\int_2^3 d\sqrt{1+x^2}$$
then in the solution the next step shows
$$\sqrt{1+x^2} |_1^2 + 2(\sqrt{1+x^2} |_2^3)$$
My question:

why is the integral of $d\sqrt{1+x^2}$ simply equal to $\sqrt{1+x^2}$?

 A: When $f'$ is continuous on $[a,b]$,
$$
\int_a^b\ df(x)=\int_a^b f'(x)\ dx=f(b)-f(a).
$$
See more information about the relationship between Riemann Stieltjes integrals and Riemann integrals here.
A: This is a generalisation of the Riemann integral:
Before we had:
$$\int_a^b dx = x|_a^b$$
Now we have
$$\int_a^b d(g(x)) = g(x)|_a^b$$
$\int_a^b d(g(x))$ is somewhat like $\int_a^b g'(x) dx$, much like in probability:
If we have a continuous random variable $X$ in $(\Omega, \mathscr F, \mathbb P)$, then
$$P(X \le x) = \int_{-\infty}^x dF_X(x) \tag{*}$$
If $X$ has a pdf, then we have
$$P(X \le x) = \int_{-\infty}^x F_X'(x) dx = \int_{-\infty}^x f_X(x) dx$$
Not all continuous random variables have pdfs so we can use $(*)$.
Same with expected value: Let $g$ be a Borel-measurable function. Then
$$E[g(X)] = \int_{\mathbb R} g(x) dF_X(x)$$
If $X$ has a pdf, then we have
$$E[g(X)] = \int_{\mathbb R} g(x) F_X'(x) dx = \int_{\mathbb R} g(x) f_X(x) dx$$
A: Owing to the uncertain integrals property
$$\int dF(x) = F(x) + const$$
