In your very last integral, $\mathrm dx$ should be $2\mathrm dx$ $(*)$. Apart from that, your reasoning looks fine. To compute the last integral, use $\displaystyle\int(2−1/x)\mathrm dx=2x−\log x$, which yields the numerical value $2−2\log2\approx61\%$ for the probability of no quadrilateral.
$(*)$ To understand why, note that the total mass of $\mathrm dx$ on $[1/2,1]$ is $1/2$ and that one wants a total mass of $1$. This indicates that the probability density function of $X$ the longest of the two first lengthes is $2$ on the interval $[1/2,1]$. Once $X=x$ is known, the probability of no quadrilateral is $p(x)=2(x-1/2)/x$, as you aptly showed, hence the total probability of no quadrilateral is
$$
\int_{1/2}^1p(x)\,(2\mathrm dx)=2\int_{1/2}^1(2-1/x)\mathrm dx.
$$
Best Answer
I have solved this puzzle earlier. So I understand what you mean. This is what you want.![enter image description here](https://i.stack.imgur.com/FoSKr.jpg)