We can prove that sequences of right- or left-hand Riemann sums will converge for a monotone function with a convergent improper integral.
Suppose WLOG $f:(0,1] \to \mathbb{R}$ is nonnegative and decreasing. Suppose further that there is a singularity at $x =0$ but $f$ is Riemann integrable on $[c,1]$ for $c > 0$ and the improper integral is convergent:
$$\lim_{c \to 0+}\int_c^1 f(x) \, dx = \int_0^1 f(x) \, dx.$$
Take a uniform partition $P_n = (0,1/n, 2/n, \ldots, (n-1)/n,1).$ Since $f$ is decreasing we have
$$\frac1{n}f\left(\frac{k}{n}\right) \geqslant \int_{k/n}^{(k+1)/n}f(x) \, dx \geqslant \frac1{n}f\left(\frac{k+1}{n}\right), $$
and summing over $k = 1,2, \ldots, n-1$
$$\frac1{n}\sum_{k=1}^{n-1}f\left(\frac{k}{n}\right) \geqslant \int_{1/n}^{1}f(x) \, dx \geqslant \frac1{n}\sum_{k=2}^nf\left(\frac{k}{n}\right). $$
Hence,
$$ \int_{1/n}^{1}f(x) \, dx +\frac{1}{n}f(1) \leqslant \frac1{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right) \leqslant \int_{1/n}^{1}f(x) \, dx+ \frac{1}{n}f \left(\frac{1}{n} \right).$$
Note that as $n \to \infty$ we have $f(1) /n \to 0$ and since the improper integral is convergent,
$$\lim_{n \to \infty} \int_{1/n}^{1}f(x) \, dx = \int_0^1 f(x) \, dx, \\ \lim_{n \to \infty}\frac{1}{n}f \left(\frac{1}{n} \right) = 0.$$
The second limit follows from monotonicity and the Cauchy criterion which implies that for any $\epsilon > 0$ and all $n$ sufficiently large
$$0 \leqslant \frac{1}{n}f \left(\frac{1}{n} \right) \leqslant 2\int_{1/2n}^{1/n}f(x) \, dx < \epsilon.$$
By the squeeze theorem we have
$$\lim_{n \to \infty}\frac1{n}\sum_{k=1}^nf\left(\frac{k}{n}\right) = \int_0^1 f(x) \, dx.$$
This proof can be generalized for non-uniform partitions. For oscillatory functions like $g(x) = \sin(1/x)/x$, the failure of the sequence of right-hand Riemann sums to converge is, non-monotonicity notwithstanding, related to non-convergence as $n \to \infty$ of
$$ \frac{1}{n}g \left(\frac{1}{n} \right) = \sin n. $$
This particular case appears to have been covered nicely by @Daniel Fischer in
Improper integrals and right-hand Riemann sums
The integral that you have written does not equal the limit when you use Reimann Sums, I believe your book is incorrect?
The middle integral is:
$$
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x
$$
which means you can't factor out a $\sqrt{3}$. So I'll answer the question using the correct Reimann Sum.
Using the Left point method:
$$
A_\mathrm{left} = \Delta x \left[f\left(a\right) + f\left(a + \Delta x\right) + f\left(a+ 2*\Delta x\right) + \cdots + f\left(b - \Delta x\right)\right]
$$
$n=\#\ \text{steps}$, $\Delta x = \frac{b-a}{n} = \frac{1}{n}$.
We want to write it in terms of steps rather than step size, so:
$$
\begin{align}
A_\mathrm{left} &= \frac{1}{n} \left[f\left(0\right) + f\left(\frac{1}{n}\right) + f\left(2*\frac{1}{n}\right) + \cdots + f\left(1 - \frac{1}{n}\right)\right] \newline
&= \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right]
\end{align}
$$
Therefore:
$$
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x = \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right]
$$
Which is not really in the nicest form, but I hope this helps!
UPDATE:
Ok so I think I misunderstood the question. So the first limit and last integral are equal to $2$, but the middle integral evaluates to $\frac{2}{3}$ which is not the same. I did the working for the middle integral thinking you were trying to determine the Reimann sum for that, but your initial question had a mistake so rather than deleting everything I will leave the working here in case anyone else finds it helpful, and answer the rest of your question.
Middle integral:
$$
\begin{align}
\int_0^1 \frac{1}{\sqrt{1+3x}}\text{d}x &= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \frac{1}{\sqrt{1+3\left(\frac{1}{n}\right)}} + \frac{1}{\sqrt{1+3\left(\frac{2}{n}\right)}} + \cdots + \frac{1}{\sqrt{1+3\left(\frac{n-1}{n}\right)}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{1}{\sqrt{1+3\left(\frac{k}{n}\right)}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} \left[1 + \sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{1}{n} + \lim_{n \rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n-1}\frac{n}{\sqrt{n^2+3kn}} \newline
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&=\frac{2}{3}
\end{align}
$$
First limit:
$$
\lim_{n \rightarrow \infty}\frac{3}{n} \sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} = 2
$$
Last integral:
$$
\sqrt{3}\int_{\frac{1}{3}}^{\frac{1}{3}+1}\frac{1}{\sqrt{x}}\text{d}{x} = 2
$$
So its clear that the middle and last integral are not the same $2\neq\frac{2}{3}$. But back to your question: Yes certain Reimann Sums may come from multiple integrals.
E.g. If the middle integrals integrand was: $3*f\left(x\right)$, then we would have:
$$
\boxed{3}*\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{n+3k}}
$$
Now we use algebra to see if we can get this to the same form as the first limit.
First let the summation index start at $k=0$ (and subtract the $k_0$th term so we don't change the expression):
$$
\begin{align}
&=\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\left[\left(\sum_{\boxed{k=0}}^{n-1}\frac{1}{\sqrt{n+3k}}\right)-\frac{1}{\sqrt{n}}\right] \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}}-\lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\frac{1}{\sqrt{n}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{\sqrt{n}}\boxed{\frac{\sqrt{n}}{\sqrt{n}}}\sum_{k=0}^{n-1}\frac{1}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n+3k}} \newline
&= \lim_{n \rightarrow \infty} \frac{3}{n}\sum_{k=0}^{n-1}\sqrt{\frac{n}{n+3k}} \newline
&= 2
\end{align}
$$
As you can see just by modifying the middle integral I was able to reach the first limit!
So I'm going to leave this here and hopefully you can see by this example that you can indeed have multiple integrals linked to a Reiman Sum.
Best Answer
We have
\begin{align*} \int_{0}^{\infty} \log \left( 1 + \frac{1}{x^2} \right) \, \mathrm{d}x &= \lim_{n\to\infty} \sum_{k=1}^{\infty} \log \left( 1 + \frac{n^2}{k^2} \right) \frac{1}{n} \\ &= \lim_{n\to\infty} \frac{1}{n} \log \left[ \prod_{k=1}^{\infty} \left( 1 + \frac{n^2}{k^2} \right) \right] \\ &= \lim_{n\to\infty} \frac{1}{n} \log \left( \frac{\sinh(n\pi)}{n\pi} \right). \end{align*}
So, the equality $\int_{0}^{\infty} \log \left( 1 + \frac{1}{x^2} \right) \, \mathrm{d}x = \pi$ is equivalent to
$$ \lim_{n\to\infty} \left( \frac{\sinh(n\pi)}{n\pi} \right)^{\frac{1}{n}} = e^{\pi}, $$
which is in fact much easier to prove only using calculus.