<s>[INST] 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
 [/INST] 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, $1.42 < I < 1.75$.
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