Change of variables and Jacobian in double integral I am supposed to evaluate a double integral using change of variables. The integral is: $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}dydx $
I am given these equations to switch between variables:
$ u=\frac{y}{x}$ and $v=x+y$
Calculations:
The first thing to calculate is the Jacobian:
$dxdy$ = $\begin{bmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ 
 \frac{\partial v}{\partial x}& \frac{\partial v}{\partial y}
\end{bmatrix}$
$\frac{\partial v}{\partial x}=\frac{\partial v}{\partial y}=1$
$\frac{\partial u}{\partial x}=\frac{-y}{x^{2}}$ and $\frac{\partial u}{\partial y}=\frac{1}{x}$
Therefore the Jacobian = $dxdy$ = $\frac{-y}{x^{2}}-\frac{1}{x}$
=  $\frac{-y-x}{x^{2}}$
and the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(x+y)e^{(x+y)}}{{x^{2}}}(\frac{-y-x}{x^{2}}) dudv $
Since $v=x+y$, 
the integral becomes $ \int_{0}^{1}\int_{0}^{x}\frac{(v)e^{(v)}(-v)}{{x^{2}}}dudv $
$ \int_{0}^{1}\int_{0}^{x}\frac{(-v^{2})e^{(v)}}{{x^{2}}}dudv $
At this point I am confused how to substitute for the x in the integral bound and the denominator as well as changing the boundaries in the integrals. I know I want the entire integral in terms of u and v. I was thinking about using $x=v-y$ but that leaves it in terms I don't want.
The answer is $e^{2}-e-1$
Thanks for the help.
 A: 1.) Transform differentials:
$$\begin{gathered}
  (x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx \hfill \\
  u = \frac{y}{x},v = x + y \hfill \\
  du =  - \frac{y}{{{x^2}}}dx + \frac{1}{x}dy \hfill \\
  dv = dx + dy \hfill \\
  du \wedge dv =  - \frac{y}{{{x^2}}}dx \wedge dy + \frac{1}{x}dy \wedge dx \hfill \\
  du \wedge dv =  - \frac{{x + y}}{{{x^2}}}dx \wedge dy = \frac{{x + y}}{{{x^2}}}dy \wedge dx \hfill \\
  \frac{1}{v}du \wedge dv = \frac{1}{{{x^2}}}dy \wedge dx \hfill \\
  \frac{1}{{{x^2}}}dydx = \frac{1}{v}dudv \hfill \\
  (x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = {e^v}dudv \hfill \\ 
\end{gathered}$$
2.) Find inverse function:
$$\begin{gathered}
  u = \frac{y}{x},v = x + y \Rightarrow uv - y = x\frac{{{y^2}}}{{{x^2}}} = x \cdot {u^2} \hfill \\
   \Rightarrow uv - x - y = x \cdot {u^2} - x \hfill \\
   \Rightarrow uv - v = x \cdot ({u^2} - 1) \hfill \\
   \Rightarrow x = \frac{{v(u - 1)}}{{{u^2} - 1}} = \frac{v}{{u + 1}} \hfill \\
  v = x + y \Rightarrow  \hfill \\
  y = v - x = v - \frac{v}{{u + 1}} = v(1 - \frac{1}{{u + 1}}) = v \cdot \frac{u}{{u + 1}} \hfill \\
  x = v \cdot \frac{1}{{u + 1}} \hfill \\
  y = v \cdot u\frac{1}{{u + 1}} \hfill \\ 
\end{gathered} $$
3.) Transform boundaries:
$$\begin{gathered}
  x = 0 \Rightarrow v = 0 \hfill \\
  x = 1 \Rightarrow v = u + 1 \hfill \\
  y = 0 \Rightarrow v \cdot u = 0 \Rightarrow u = 0 \vee v = 0 \hfill \\
  y = x \Rightarrow u = 1 \hfill \\ 
\end{gathered}$$
4.) Integrate:
$$\begin{gathered}
  \int\limits_0^1 {\int\limits_0^x {(x + y){e^{x + y}}\frac{1}{{{x^2}}}dydx = \int\limits_0^1 {\int\limits_0^{u + 1} {{e^v}dvdu} } } }  = \int\limits_0^1 {({e^{u + 1}} - 1)du}  \hfill \\
  \int\limits_0^1 {({e^{u + 1}} - 1)du}  = {e^2} - e - 1 \hfill \\ 
\end{gathered}$$
A: $ \int_{0}^{1}\int_{0}^{x}\frac{(-v^{2})e^{(v)}}{{x^{2}}}dudv $
$ y=u x,  v=x + u x= (1+u)x,  x=v/(1+u) , $  plug in for x
$ \int_{0}^{1}\int_{0}^{x} -e^v (1+u^2) du dv $
