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
 A: $$\int_0^1\sqrt{1+\dfrac1{3x}}dx=2\int_0^1\sqrt{t^2+\dfrac13}dt$$ proves convergence.
Then
$$\frac1{\sqrt 3}\le\sqrt{t^2+\frac13}\le t+\frac1{\sqrt3}$$
implies
$$\frac2{\sqrt 3}\approx 1.155\le I\le1+\frac2{\sqrt 3}\approx2.155$$
A tighter upper bound is obtained by noting that the function is convex and
$$\sqrt{t^2+\frac13}\le \frac1{\sqrt3}+t\left(\sqrt{\frac 43}-\frac1{\sqrt3}\right),$$ giving $$I\le\sqrt3\approx1.732$$
A tighter lower bound could be found by considering the tangents at both endpoints up to their intersection, but we can already conclude C.

The exact value is $$1.5936865\cdots$$ The bounds can be computed by hand, by squaring to avoid square roots.
A: Starting from
$$\int_0^1\sqrt{1+{1\over3x}}\,dx=2\int_0^1\sqrt{t^2+{1\over3}}\,dt$$
(from the subsitution $x=t^2$) as in Yves Daoust's answer, integration by parts gives
$$\int_0^1\sqrt{t^2+{1\over3}}\,dt=t\sqrt{t^2+{1\over3}}\Big|_0^1-\int_0^1{t^2\over\sqrt{t^2+{1\over3}}}\,dt={2\over\sqrt3}-\int_0^1{t^2+{1\over3}-{1\over3}\over\sqrt{t^2+{1\over3}}}\,dt$$
hence
$$2\int_0^1\sqrt{t^2+{1\over3}}\,dt={2\over\sqrt3}+{1\over3}\int_0^1{dt\over\sqrt{t^2+{1\over3}}}={2\over\sqrt3}+{1\over\sqrt3}\int_0^1{dt\over\sqrt{3t^2+1}}$$
Since $1\le\sqrt{3t^2+1}\le2$ for $0\le t\le1$, we have
$${1\over2}\le\int_0^1{dt\over\sqrt{3t^2+1}}\le1$$
Thus
$${2\over\sqrt3}+{1\over2\sqrt3}\le2\int_0^1\sqrt{t^2+{1\over3}}\,dt\le{2\over\sqrt3}+{1\over\sqrt3}$$
Now
$${2\over\sqrt3}+{1\over2\sqrt3}={5\sqrt3\over6}=\sqrt{75\over36}\gt\sqrt2\gt1.4$$
and
$${2\over\sqrt3}+{1\over\sqrt3}=\sqrt3\lt\sqrt{3.24}=1.8$$
Consquently
$$1.4\lt\int_0^1\sqrt{1+{1\over3x}}\,dx\lt1.8$$
and thus (C) is the correct answer.
A: Since $\sqrt{t^2+1/3}$ is a convex function on $[0,1]$, you may simply use the Hermite-Hadamard inequality to derive that
$$ \sqrt{2+\frac{1}{3}}\leq 2\int_{0}^{1}\sqrt{t^2+1/3}\,dt \leq \sqrt{3} $$
so $(C)$ is the correct option.
A: 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  $(C)$
A: Let's try to use integration by parts to $I = \int\limits_0^1 \sqrt{1 + \frac{1}{3x}}dx$. First, transform integral into $\frac{2}{\sqrt3}\int\limits_0^1\frac{\sqrt{1 + 3x}}{2\sqrt{x}}$. Now $u = \sqrt{3x+1}$ and $dv = \frac{dx}{2\sqrt{x}}$ and what we get after IBP is $$\frac{2}{\sqrt3}\sqrt{x(3x+1)}|_0^1 - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x+1}}dx$$. We have $$\frac{5}{2\sqrt3} = \frac{4}{\sqrt3} - \sqrt{3}\int\limits_0^1 \sqrt{\frac{x}{3x + x}}dx < I <\frac{4}{\sqrt3} - \sqrt3\int\limits_0^1 \sqrt{\frac{x}{3 + 1}}dx = \frac{4}{\sqrt3} - \frac{\sqrt3}{2} \frac{2}{3}x\sqrt{x}|_0^1 = \sqrt3$$
$\frac{5}{2\sqrt3} \approx 1.44$ and $\sqrt3 \approx 1.73$, so the answer is (C).
If one doesn't know the value of $\sqrt3$, we can check that $1.7^2 < 3 < 1.8^2$ and then $3 < 1.75^2$. Therefore, $\sqrt3 < 1.75$. From it we have $\frac{5}{2\sqrt3} > \frac{5}{2\cdot1.75} > 1.42$ and, for integral, $1.42 < I < 1.75$.
