UPDATE: Apparently a bounded region with 3 sides can expand to a bounded region with 4 sides after a transformation. I was not aware of this. Though it seems like a reasonable thing to happen.
I've been working on this problem for about 6 hours.
$ I = \int^1_0dx\int^x_0\frac{(x+y)e^\frac{x+y}{x-y}}{x^2}dy$
with the change of variables:
$ u = y/x$
$ v = x+y$
I calculated the jacobian determinant to be:
$|J| = \frac{v}{1+u^2}$
alot of stuff cancels nicely since,
$x^2 = \frac{v^2}{(1+u)^2}$
then I just becomes a simple integral over the new region Q
$ I = \int\int_Qe^vdvdu$
To get this region I noted that the integral region in x,y is a triangle which is the intersection of the following lines:
$x=1$
$y=0$
$x=y$
which correspond to the triangular region Q of u-v plane given by the intersection of:
$ u-v = -1$
$ v = 0 $
$ u = 1 $
However pluging these boundaries into the integral gives:
$ I = \int^2_0e^vdv\int^1_{v-1}du$
This clearly evaluates to $I = e^2 -3$ whereas the correct answer from the back of the textbook is $ I = e^2-e-1$
this is kind of drivin' me nuts, would appreciate any insight as to what i might be doing wrong.