Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$. I'm trying to solve this:

Which of the following is the closest to the value of this integral?
$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge.

I've found a lower bound by manually calculating $\int_{0}^{1} \sqrt{1+\frac{1}{3}} \ dx \approx 1.1547$. This eliminates option (A). I also see no reason why the integral shouldn't converge. However, to pick an option out of (B), (C) and (D) I need to find an upper bound too. Ideas? Please note that I'm not supposed to use a calculator to solve this.
From GRE problem sets by UChicago

We know that,
${\displaystyle\int}\sqrt{\dfrac{1}{3x}+1}\,\mathrm{d}x$=$=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{\sqrt{3}}}}{\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$
Substitute $u=\sqrt{\dfrac{1}{x}+3}$ and $\dfrac{\mathrm{d}u}{\mathrm{d}x} = -\dfrac{1}{2\sqrt{\frac{1}{x}+3}x^2}$,i.e $\mathrm{d}x=-2\sqrt{\dfrac{1}{x}+3}x^2\,\mathrm{d}u$
${\displaystyle\int}\sqrt{\dfrac{1}{x}+3}\,\mathrm{d}x$=$-\class{steps-node}{\cssId{steps-node-2}{2}}{\displaystyle\int}\dfrac{u^2}{\left(u^2-3\right)^2}\,\mathrm{d}u$
$={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-5}{u^2-3}}}{\left(u^2-3\right)^2}+\dfrac{\class{steps-node}{\cssId{steps-node-6}{3}}}{\left(u^2-3\right)^2}\right)\mathrm{d}u$
$={\displaystyle\int}\dfrac{1}{u^2-3}\,\mathrm{d}u+\class{steps-node}{\cssId{steps-node-7}{3}}{\displaystyle\int}\dfrac{1}{\left(u^2-3\right)^2}\,\mathrm{d}u$
Perform partial fraction decomposition:
$={\displaystyle\int}\left(\dfrac{1}{2\sqrt{3}\left(u-\sqrt{3}\right)}-\dfrac{1}{2\sqrt{3}\left(u+\sqrt{3}\right)}\right)\mathrm{d}u$ + ${\displaystyle\int}\dfrac{1}{u+\sqrt{3}}\,\mathrm{d}u$
On solving this further we get
$\dfrac{\sqrt{\frac{1}{3x}+1}}{3\left(\frac{1}{3x}+1\right)-3}+\dfrac{\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)}{6}-\dfrac{\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$
I.e,
$\dfrac{6\sqrt{\frac{1}{3x}+1}x+\ln\left(\sqrt{\frac{1}{x}+3}+\sqrt{3}\right)-\ln\left(\left|\sqrt{\frac{1}{x}+3}-\sqrt{3}\right|\right)}{6}+C$
$\boldsymbol{\int\limits^{1}_{0}{f(x)}\,\mathrm{d}x =}$=${1\over6}[4 \sqrt(3) + \ln(7 + 4 \sqrt(3)]$
Approximation:
$1.593686504020857$,
i.e $1.6$.
The answer is option  $(C)$
