Substitution in a double integral I have trouble in doing a substitution. That is one step in a whole demonstration. I need to compute this integral : $\epsilon << 1 $
$$  \\  \int_{ [0,1-\epsilon] } \int_{ [0,1-\epsilon] } \frac 1 {1-xy} \, dx dy$$ 
And for that I need to do the substitution : 
$$ x = u-v \, ; \, y = u + v $$
Can you help me to do it? I'm not asking for the whole calculus, I have the answer under the eyes. My problem is that it is one of the first multi-integral I'm computing and I struggle to understand what is depending of what, what do I need to write like a function of the rest etc...  
For example, 


*

*I know that I need to compute the determinant so this I ve done : it
gives 2. 

*I know that I need to find the lines that bound my
submanifold, and so I have written :


$$
\left\{ 
\begin{array}{c}
y = 0 \\ 
x = 0 \\ 
y = 1 - \epsilon \\
x = 1 - \epsilon
\end{array}
\right. 
 \implies 
\left\{ 
\begin{array}{c}
u= -v \\ 
u = v \\ 
u+v = 1 - \epsilon \\
u-v = 1 - \epsilon
\end{array}
\right. 
$$


*

*I know that I need to find the boundaries for u and v :
$$
\left\{ 
\begin{array}{c}
0 < x < 1 - \epsilon \\
0 < y < 1 - \epsilon 
\end{array}
\right. 
 \implies 
\left\{ 
\begin{array}{c}
0 < u < 1 - \epsilon \\
-1 + \epsilon < v < 1 - \epsilon
\end{array}
\right. 
$$


I would be so greatfull if you can help me figure out what I need to  do in order to have a general method to deal with problems like this. 
thank you!
P.S. : please, so many times, people told me just to do a graph... I have done it !!!! I just need some help at first, and I'm so sorry guys to ask you this, but it's really necessary for me... I tried for 2 days now :(

On the $uv$-plane, we can graph the intersection of the following inequalities
$$
\begin{align}
u-v > 0 \\
u-v < 1 - \epsilon \\
u+v > 0 \\
u+v < 1 - \epsilon
\end{align}
$$
Here is what the intersection looks like.
When $u \in \left[0, \dfrac{1-\epsilon}{2}\right]$, it follows then that $v \in [-u,u]$. Otherwise, when $u \in \left[\dfrac{1-\epsilon}{2}, 1-\epsilon\right]$, it follows that $v \in [(u+\epsilon) - 1, 1 - (u+\epsilon)]$. Setting up the integrals:
$$I = \int_{0}^{(1-\epsilon)/2} \int_{-u}^{u} \frac{2}{1-(u^2-v^2)} \,  dvdu + \int_{(1-\epsilon)/2}^{1-\epsilon} \int_{(u+\epsilon)-1}^{1-(u+\epsilon)} \frac{2}{1-(u^2-v^2)} \,  dvdu$$
