 Solution

Take the usual positive, counterclockwise boundary orientation; the printed question does not specify one, and reversing it negates the answer. Write the successive corner points as $P=(1,1)$, $Q=(\sqrt{2},1/\sqrt{2})$, $R=(2,1)$, $S=(\sqrt{2},\sqrt{2})$. Traverse $P\to Q\to R\to S\to P$, keeping the region on the left.

On $PQ$, use $y=1/x$ with $1\le x\le \sqrt{2}$. Here $dy=-dx/x^2$, so the line integral isOn $QR$, use $y=x/2$ with $\sqrt{2}\le x\le 2$:On $RS$, use $y=2/x$ with $x$ decreasing from $2$ to $\sqrt{2}$:Finally, on $SP$, use $y=x$ with $x$ decreasing from $\sqrt{2}$ to $1$:Adding all four line integrals gives ${\oint }_C(x^2/(2y))dy-dx=1/2$ for the positive orientation, or $-1/2$ for the reverse orientation.