Answer
$$\iint_{R}\left(y+x y^{-2}\right) d A =4$$
Work Step by Step
Given $$\iint_{R}\left(y+x y^{-2}\right) d A, \quad R=\{(x, y) | 0 \leqslant x \leqslant 2,1 \leqslant y \leqslant 2\}$$
So, we have
\begin{aligned}I&= \iint \limits_{R}\left(y+x y^{-2}\right) d A\\
&=\int_{0}^{2} \int_{1}^{2}\left(y+x y^{-2}\right) d y d x\\
&=\int_{0}^{2}\left[\frac{y^{2}}{2}-x y^{-1}\right]_{y=1}^{y=2} d x\\
&=\int_{0}^{2}\left[ \frac{4}{2}-\frac{x}{2}-\frac{1}{2}+x\right] d x\\
&=\int_{0}^{2}\left[ \frac{4}{2}+\frac{x}{2}-\frac{1}{2} \right] d x\\
&= \left[ \frac{3x}{2}+\frac{x^2}{4 }\right]_{0}^{2} \\
&=\left[ \frac{6}{2}+\frac{2^2}{4}\right]-\left[ 0+\frac{0^2}{4}\right] \\
&=4
\end{aligned}
