Answer
$4$
Work Step by Step
$\displaystyle \iint_{R}(y+xy^{-2})dA=\int_{1}^{2}\int_{0}^{2}(y+xy^{-2})dxdy=\int_{1}^{2}[\int_{0}^{2}(y+xy^{-2})dx]dy$
$\left[\begin{array}{l}
\text{... treat y as a constant in the inner integral} \\
\displaystyle\int_{0}^{2}(y+xy^{-2})dx=[xy+\frac{y^{-2}}{2}x^{2}]_{x=0}^{x=2}=\\ \\
=2y+\dfrac{4y^{-2}}{2}\\\\
=2y+2y^{-2}
\end{array}\right]$
$=\displaystyle \int_{1}^{2}(2y+2y^{-2})dy$
$=[y^{2}-2y^{-1}]_{1}^{2}$
$=(4-1)-(1-2)$
$=4$
