Double Integral Change of variable help I am having some trouble getting this problem set up, and would appreciate any help.
Problem: $\iint \frac{1}{(x+y)^2} dA$.  Integrate using change of variables over the region inside the lines $x+y=1,\; x+y=4$ and the $x$ and $y$ axes.
 I start off setting $u = x+y$, so $1\leq u\leq4$ . But I can't for the life of me figure out what to set as $v$ (second variable).
 A: Let $u= (x+y)$ and $v= (x-y)$ whereupon inverting we have $x=\frac12 (u+v)$ and $y=\frac12 (u-v)$.
Then, the transformed region in the $u-v$ plane is bounded as follows.  
The line $x+y=1$ becomes $u=1$.
The line $x+y=4$ becomes $u=4$.
The line $x=0$ becomes $u=-v$.
The line $y=0$ becomes $u=v$.
The absolute value to the Jacbobian is trivially seen to be $\frac12$.
Finally, 
$$\begin{align}\int \int \frac{1}{(x+y)^2}dxdy&=\int_1^4 \int_{-u}^{u} \frac{1}{u^2}\frac12 dvdu\\\\
&=\log 4
\end{align}$$

We can obtain the identical answer using the original coordinate system.  
We will split the region of integration into two pieces.
Region 1 extends from $0$ to $1$ on $x$ and from $1-x$ to $4-x$ on $y$.
Region 2 extends from $1$ to $4$ on $x$ and from $0$ to $4-x$ on $y$.
Then, we have 
$$\begin{align}\int \int \frac{1}{(x+y)^2}dxdy&=\int_0^1 \int_{1-x}^{4-x} \frac{1}{x^2+y^2} dydx+\int_1^4 \int_{0}^{4-x} \frac{1}{x^2+y^2} dydx\\\\
&=\frac34+(\log 4-\frac34)\\\\
&=\log 4
\end{align}$$
as expected!
