Find area bounded by $y=\frac 3x, y=\frac 5x, y=3x, y=6x$ Let S be the area of the region bounded by the curves 
$$y=\frac 3x,\>\>\> y=\frac 5x,\>\>\> y=3x,\>\>\> y=6x$$
Need to find $S$.
The coordinates of the vertices of the resulting figure were found. The problem with the transition from a double integral to a repeated one.
There's another idea. Make a replacement
\begin{cases}
\xi=xy \\ \eta=\frac{y}{x}
\end{cases}
Вut none led to the correct answer.

1) Subtract the two integrals of $5/x$ and $3/x$ in the range of $a$ which is the solution of $\frac{3}{x}=6x$ (the intersection of two lines) to $b$ which is the solution of $\frac{5}{x}=3x$.
2) One would obtain $area=[ln(2x)]_{a}^{b}$.
3)Now, to subtract the area of the two hyperbolic triangles, simply obtain $c$, the solution of $\frac{5}{x}=6x$ and $d$, the solution of $\frac{3}{x}=3x$ and integrate the function $5/x$ from a to c minus the integral of function $6x$ over the same range. Do the same for the other two terms ($3/x$ and $3x$) over range $b$ to $d$. (${a,b,c,d}\in R^+$)
$a=\frac{\sqrt{2}}{2},b=\sqrt{\frac{5}{3}},c=\sqrt{\frac{5}{6}},d=1$
Now, $S = area - (\int_{a}^{c}\frac{5}{x} - 6x dx ) - (\int_{b}^{d}\frac{3}{x} - 3x dx ) = [ln(2x)]_{a}^{b} - [ln(5x)-3x^2]_{a}^{c}-[ln(3x)-1.5x^2]_{b}^{d}$
Since $\ln(a)+\ln(b)=\ln(ab), \ln(a)-\ln(b)=\ln(a/b)$ then the above can be written as $\ln(\frac{b}{cd})+3c^{2}-3a^{2}+1.5d^{2}-1.5b^{2}=\ln(\sqrt{2})$
Since it is a hyperbola (mirror image on 4th quadrant), multiply the area between the two functions $5/x$ and $3/x$ by two. So, $2\ln(\sqrt{2})=\ln2$.
