
Our region (highlighted in yellow) is bound by curves $x^2 - y^2 = 1, x^2 - y^2 = 9, xy = 2$ and $xy = 4$ in the first octant as ($x \geq 0, y \geq 0$).
Equating $x^2 - y^2 = 1, xy = 2$,
$(\frac{2}{y})^2 - y^2 = 1 \implies y^4 + y^2 = 4$. Consider this as a quadratic in $y^2$ and solve for $y^2$. We find $y \approx 1.25$ and so $x = \frac{2}{y} = \approx 1.6$.
Similarly find intersection with $xy = 4$.
Then find intersection of $x^2-y^2 = 9$ with $xy = 2$ and $xy = 4$.
We obtain intersection points as $(1.6, 1.25), (2.13, 1.88), (3.08, 0.65)$ and $(3.25, 1.23)$.
Keeping things simple, I will set up the integral in $3$ parts (please see sketch) -
i) $1.6 \leq x \leq 2.13$ between curves $xy \geq 2$ and $x^2-y^2 \geq 1$
ii) $2.13 \leq x \leq 3.08$ and $2 \leq xy \leq 4$
iii) $3.08 \leq x \leq 3.25$ between $x^2-y^2 \leq 9$ and $xy \leq 4$.
$\displaystyle I_1 = \int_{1.6}^{2.13}\int_{2/x}^{\sqrt{x^2-1}} (x^2+y^2) dy dx \approx 1.48$
$\displaystyle I_2 = \int_{2.13}^{3.08}\int_{2/x}^{4/x} (x^2+y^2) dy dx \approx 6.02$
$\displaystyle I_3 = \int_{3.08}^{3.25}\int_{\sqrt{x^2-9}}^{4/x} (x^2+y^2) dy dx \approx 0.5$
$I = I_1 + I_2 + I_3 \approx \fbox{8}$