Smaller over larger

Idea. Rather than integrating the ratio directly, find its distribution. It turns out to be uniform, and then the expectation is immediate.

Step 1: reduce to one order. The events $\{U<V\}$ and $\{V<U\}$ are symmetric and together have probability $1$. On $\{U<V\}$, the ratio is $R=U/V$. So for $0\le r\le 1$, $$\mathbb{P}(R\le r)=2\mathbb{P}(U\le rV).$$ (When $U\le rV$ with $r\le 1$, automatically $U\le V$.)

Step 2: an area. The region $\{(u,v):0\le u\le rv,0\le v\le 1\}$ is a triangle with vertices $(0,0)$, $(0,1)$ and $(r,1)$. Its area is $\frac{r}{2}$. So $$\mathbb{P}(R\le r)=2\cdot \frac{r}{2}=r.$$

Step 3. $R$ is uniform on $[0,1]$, so $\mathbb{E}[R]=\frac{1}{2}$.

Remark. This is a scale property: given the larger value $V=v$, the smaller one is uniform on $[0,v]$, so the ratio is uniform on $[0,1]$ whatever $v$ is.