$$ \iint_D\frac{x-y+1}{x+y}\,d(x,y)$$
$$D: 1 \leq x+y \leq 2x-4y \leq 3$$
I was trying to solve this integral using uv substitution. I called: $$u = x+y $$ And: $$v = 2x-4y$$
I can find the Jacobian and from there the determinant, however, I get stuck on the actual substitution. I do know that the denominator can be put to $u,$ but what do I change the numerator to? Or do I have the wrong substitution for $u$ and $v$?
If we take $$u=x+y\\v=2x-4y$$ then we get $$x=\frac{4u+v}{6} \\ y=\frac{2u-v}{6}$$ Note that $\frac{\partial(x,y)}{\partial(u,v)}=-\frac{1}{6}$ and the image of your region $D$ under $(x,y)\longrightarrow (x+y,2x-4y)$ is the region $\tilde{D}$ defined by $$\tilde{D}:1\leq u \leq v \leq 3$$ So we get $$\iint_{D}\frac{x-y+1}{x+y}\mathrm{d}y\mathrm{d}x=\frac{1}{6}\int_1^3 \int_u^3\Bigg[\frac{v}{3u}+\frac{1}{u}+\frac{1}{3}\Bigg]\mathrm{d}v\mathrm{d}u$$ Can you finish?