Double integral over region bounded by nonparallel lines Problem
I'm having trouble with the following integral:
$\int \int_D \frac{x}{y} dx dy $  where D is the area bounded by $1 \leq 2x+y \leq 5, \quad 4x \leq y \leq 8x$
My attempt at a solution
Substitution: 
$ u = 2x + y, \quad v= y \ \implies x = \frac{1}{2}(u-v)$
Jacobian: 
$ \begin{vmatrix} \frac{1}{2} & -\frac{1}{2} \\ 0 & 1 \end{vmatrix} = \frac{1}{2}$
Limits for u:
$u: 1\to 5$
Limits for y:
$4x \leq y \implies 2(u-v) \leq v \implies \frac{2}{3}u \leq v \\
y \leq 8x \implies v \leq 4(u-v) \implies v \leq \frac{4}{5}u$
$v: \frac{2}{3}u \to \frac{4}{5}u$
Everything put together:
$\int \int_D \frac{x}{y} dx dy = \int \int_D \frac{\frac{1}{2}(u-v)}{v} \cdot \frac{1}{2} dudv = \int_1^5 \int_{\frac{2}{3}u}^{\frac{4}{5}u} \Big(\frac{u}{4v}-\frac{1}{4} \Big)dv du$
I put this into wolframs double integral calculator and get:
$3 \log(\frac{6}{5})-\frac{2}{5}$ which is wrong. Am I doing something wrong?

Hint. Integrate w.r.t. $y$ immediately:
$$\int_{4x}^{8x} \frac{dy}{y}=\log 2$$
Now the only thing you have left is finding the limits for $x$ from the two inequalities and integrating:
$$\log 2\int_{x_1}^{x_2} xdx$$

Note that the first integration is equivalent to a change of variable:
$$y=x t$$
$$\int_{4x}^{8x} \frac{dy}{y}=\int_{4}^{8} \frac{dt}{t}$$
