I have to integrate this: $$ \int_{0}^{1}\int_{0}^{1 - x} \exp\left(\frac{y}{x + y}\right) \mathrm{d}y\,\mathrm{d}x $$ using the substitution $x + y = u$ and $y = uv$.
Now, under the given transformation boundary
$x=0$ changes to $v=1$;
$y=1-x$ changes to $u=1$;
$y=0$ gives $v=0$
but what about the boundary $x=1$, where does it go ?.
Consider the transformation $\Phi(u,v)=(x,y)$ given by $x=u(1-v)$ and $y=uv$. Then the Jacobian is $$\frac{\partial(x,y)}{\partial(u,v)}=u.$$ Moreover $\Phi$ is a bijection, with inverse $u = x+y$, $v = y/(x+y)$, between the open square $$Q=\{(u,v): u\in (0,1),v\in (0,1)\}$$ and the open triangle $$T=\{(x,y): x\in (0,1), 0< y< 1-x\}.$$ Therefore $$\iint_T e^{\frac{y}{x+y}} \ dy\ dx=\iint_Q e^{v}u \ du\ dv=\frac{e-1}{2}.$$
P.S. Given $v\in [0,1)$, $(x,y)=(u-uv,uv)$ for $u\in (0,1]$ is the parametric equation of a segment along the line $y=\frac{v}{1-v}\cdot x$.