How to evaluate $\int\limits^1_0 \sqrt{1+\frac{1}{x}}\, \text{d}x$

Here's something you might try. Note that the length of that arc will be the same as the length of the same arc, reflected over the line $y=x$. That is, the arc $y=x^2/4$, from $x=0$ to $x=2$.

Substitute $u=\sqrt{1+\frac{1}{x}}=\sqrt{\frac{x+1}{x}}$. Then $x=\frac{1}{u^2-1}$, so $dx=-\frac{2u}{(u^2-1)^2}du$, which makes: $$\int\sqrt{1+\frac{1}{x}}\,{dx}=-\int \frac{2u^2}{(u^2-1)^2}\,{du}$$ Can you continue from here?

A good thing to note here is that $y = 2 \sqrt x$ is the same as $x = \frac{y^2}{4}$. So by 'swapping the order of integration' (sort of), you calculate a much easier integral. But do remember that the domain for $x$ here is $[0,1]$, but for $y$ it's $[0,2]$.