Answer
$ \dfrac{8}{5} \ln 8$
Work Step by Step
$Jacobian, J(x,y) =\begin{vmatrix} u_x&u_y\\v_x&v_y\end{vmatrix}=\begin{vmatrix} 1&-2\\3& -1\end{vmatrix}=1 \cdot (-1) -(-2)(3)=5$
and $J(u,v)=|\dfrac{1}{5}|$
Now, $\iint_R \dfrac{x-2y}{3x-y}dA=\int_1^{8} \int_{0}^{4})(\dfrac{1}{5})\dfrac{u}{v} du dv$
or, $= \int_0^{4} u du \int_{1}^{8} (\dfrac{1}{5})(\dfrac{1}{v}) dv$
Therefore, $\iint_R \dfrac{x-2y}{3x-y}dA=(\dfrac{1}{5}) (\dfrac{u^2}{2})_0^4 [\ln v]_{1}^{8}= \dfrac{8}{5} \ln 8$
