Answer
$\dfrac{8}{5} \ln 8$
Work Step by Step
Plug $u=x-2y$ and $v=3x-y$
Here, we have: $Jacobian =\begin{vmatrix} u_x&u_y\\v_x&v_y\end{vmatrix}=1 \cdot (-1) -(-2)(3)=5$
Now, $\iint_R \dfrac{x-2y}{3x-y}dA=(\dfrac{1}{5}) \int_1^{8} \int_{0}^{4}uv^{-1} du dv$
This implies that
$\iint_R \dfrac{x-2y}{3x-y}dA=(\dfrac{1}{5}) \int_0^{4} u du \int_{1}^{8} v^{-1} dv=(\dfrac{1}{5}) (u^2/2)_0^4 [\ln v]_{1}^{8}$
Hence, we get$\iint_R \dfrac{x-2y}{3x-y}dA= (\dfrac{8}{5}) \ln 8$
