Let $$\mathbb{R}$$ denote the set of all real numbers. Then the area of the region

$$\left\{(x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0 \right\}$$

is

We must find the area of the set

$$S=\Bigl\{(x,y)\in\mathbb{R}\times\mathbb{R}\;:\;x\gt 0,\;y\gt\frac1x,\;5x-4y-1\gt 0,\;4x+4y-17\lt 0\Bigr\}.$$

Rewrite the straight-line inequalities in the usual $$y=m x+c$$ form:

$$5x-4y-1\gt0\;\Longrightarrow\;4y\lt5x-1\;\Longrightarrow\;y&lt\frac{5x-1}{4},$$
$$4x+4y-17&lt0;;\Longrightarrow\;4y&lt17-4x\;\Longrightarrow\;y&lt\frac{17-4x}{4}=\,\frac{17}{4}-x.$$

Therefore every point of $$S$$ must satisfy

$$y&gt\frac1x \quad\text{and}\quad y&lt\min\!\Bigl\{\frac{5x-1}{4},\,\frac{17}{4}-x\Bigr\}.$$

Call the two lines

$$\ell_1:\;y=\frac{5x-1}{4}\qquad \ell_2:\;y=\frac{17}{4}-x.$$

Intersection of the two lines
Set $$\frac{5x-1}{4}=\frac{17}{4}-x$$ to obtain

$$5x-1=17-4x\;\Longrightarrow\;9x=18\;\Longrightarrow\;x=2$$
and hence $$y=\frac{5(2)-1}{4}=\frac94.$$
Thus $$\ell_1\cap\ell_2=(2,\tfrac94).$$

Which line is lower?
Solve $$\frac{5x-1}{4}\lt\frac{17}{4}-x\quad\Longrightarrow\quad x&lt2.$$
Hence
for $$0&lt x&lt2$$ the lower (controlling) upper-boundary is $$\ell_1,$$
for $$x&gt2$$ the lower upper-boundary is $$\ell_2.$$

Feasibility with the hyperbola
The region exists only where the hyperbola is below the chosen line:

1. For $$y&lt\ell_1$$ (i.e. $$0&lt x&lt2$$) we need $$\frac1x&lt\frac{5x-1}{4}\;.$$ Multiply by $$4x\;(x&gt0):\quad4&lt5x^{2}-x\;,$$
$$5x^{2}-x-4&gt0;;\Longrightarrow\;x&gt1\quad(\text{rejected root }x&lt0).$$

2. For $$y&lt\ell_2$$ (i.e. $$x&gt2$$) we need $$\frac1x&lt\frac{17}{4}-x.$$ Multiply by $$4x\;(x&gt0):\quad17x-4x^{2}-4&gt0,$$
$$-4x^{2}+17x-4&gt0;;\Longrightarrow\;4x^{2}-17x+4&lt0.$$
This quadratic has roots $$x=\frac{17\pm15}{8}\;=\;0.25,\;4,$$ so the inequality holds for $$0.25&lt x&lt4.$$ Intersecting with $$x&gt2$$ leaves the range $$2&lt x&lt4.$

Thus the region splits into two x-intervals

• $$1&lt x&lt2:\quad$$\frac$$1x&lt y&lt$$\frac{5x-1}{4}$$,$$
• $$2&lt x&lt4:\quad$$\frac$$1x&lt y&lt$$\frac{17}{4}$$-x.$$

Area computation

Area $$A=$$\int_{1}^{2}\Bigl(\frac{5x-1}{4}-\frac1x\Bigr$$)\,dx+$$\int_{2}^{4}\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)\,dx.$$

First integral:

$$$$\int_{1}^{2}\!\Bigl(\frac{5x}{4}-\frac14-\frac1x\Bigr$$)dx =\Bigl[$$\frac{5x^{2}$$}{8}\Bigr]_{1}^{2}-\Bigl[$$\frac{x}{4}$$\Bigr]_{1}^{2}-\bigl[$$\ln$$ x\bigr]_{1}^{2}$$ $$=$$\frac{5}{8}$$(4-1)-$$\frac$$14(2-1)-$$\ln$$2 =$$\frac{15}{8}-\frac$$14-$$\ln$$2 =$$\frac{13}{8}-\ln$$2.$$

Second integral:

$$$$\int_{2}^{4}\!\Bigl(\frac{17}{4}-x-\frac1x\Bigr$$)dx =\Bigl[$$\frac{17x}{4}$$\Bigr]_{2}^{4}-\Bigl[$$\frac{x^{2}$$}{2}\Bigr]_{2}^{4}-\bigl[$$\ln$$ x\bigr]_{2}^{4}$$ $$=$$\frac{17}{4}$$(4-2)-$$\frac$$12(16-4)-($$\ln$$4-$$\ln$$2)$$ $$=$$\frac{17}{2}$$-6-$$\ln$$2 =$$\frac{5}{2}-\ln$$2.$$

Adding:

$$A=\Bigl($$\frac{13}{8}-\ln$$2\Bigr)+\Bigl($$\frac{5}{2}-\ln$$2\Bigr) =$$\frac{13}{8}+\frac{20}{8}$$-2$$\ln$$2 =$$\frac{33}{8}-\ln$$4.$$

Final result

Area $$=\displaystyle$$\frac{33}{8}-\log_e4$$.$$ Hence the correct option is

Option B which is: $$\dfrac{33}{8}-$$\log_e4$$$$.