Let $\mathbb{R}$ denote the set of all real numbers. Then the area of the region $\{(x, y) \in \mathbb{R} \times \mathbb{R} : x > 0,\, y > \frac{1}{x},\, 5x - 4y - 1 > 0,\, 4x + 4y - 17 < 0\}$ is

$x > 0,\quad y > \dfrac1x$ $5x - 4y - 1 > 0,\quad 4x + 4y - 17 < 0$ $\text{To find: area of the region}$ $L_1: y = \dfrac{5x-1}{4},\quad L_2: y = \dfrac{17-4x}{4}$ $H: y = \dfrac1x$ $\dfrac1x < y < \min\!\left(\dfrac{5x-1}{4},\ \dfrac{17-4x}{4}\right)$ $L_1 \le L_2 \iff x \le 2$ $H \cap L_1:\ x = 1,\ y = 1$ $L_1 \cap L_2:\ x = 2,\ y = \dfrac94$ $H \cap L_2:\ x = 4,\ y = \dfrac14$ $A = \int_1^2\!\left(\dfrac{5x-1}{4} - \dfrac1x\right)dx + \int_2^4\!\left(\dfrac{17-4x}{4} - \dfrac1x\right)dx$ $\int_1^2\!\left(\dfrac{5x-1}{4} - \dfrac1x\right)dx = \dfrac{13}{8} - \ln 2$ $\int_2^4\!\left(\dfrac{17-4x}{4} - \dfrac1x\right)dx = \dfrac52 - \ln 2$ $A = \dfrac{13}{8} + \dfrac52 - 2\ln 2 = \dfrac{33}{8} - \ln 4$ $\boxed{\dfrac{33}{8} - \log_e 4} \quad (2)$