Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$.

$$\int_0^1\sqrt{1+\dfrac1{3x}}dx=2\int_0^1\sqrt{t^2+\dfrac13}dt$$ 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.

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.