<s>[INST] Resolve this double integral $\iint_D x(y+x^2)e^{y^2-x^4}dxdy$ over $D=\{ x \geq 0, x^2\leq y \leq x^2+1,2-x^2\leq y \leq 3-x^2 \}$ We can rewrite 
$$
I=\iint_D x(y+x^2)e^{(y-x^2)(y+x^2)} dx dy
$$
$$
D= \{ (x,y) \in \mathbb{R}^2:x \geq 0 \land 0 \leq y-x^2 \leq 1 \land 2 \leq y+x^2 \leq 3 \}.
$$
With this new notation we can let 
$$
\Phi(x,y) = 
\begin{pmatrix}
  y+x^2 \\
  y-x^2
\end{pmatrix}
=
\begin{pmatrix}
  u \\
  v
\end{pmatrix}
$$
 and
$$
J_{\Phi}(x,y) = 
  \begin{vmatrix}
    2x & 1 \\
   -2x & 1
  \end{vmatrix}
=
2x +2x = 4x.
$$
Hence
$$\begin{align*} 
I &= \int_2^3 \left( \int_0^1 x u e^{uv} \frac{1}{ |4x| } dv \right) du \\  
  &= \frac{1}{4} \int_2^3 \left( u  \int_0^1 e^{uv} dv \right) du \\
  &= \frac{1}{4} \int_2^3 \left( u \int_0^1 e^{uv}  dv \right) du \\
  &= \frac{1}{4} \int_2^3 \left( u \frac{1}{u} e^{uv} \Big\vert_{v=0}^{v=1}  \right) du \\
  &= \frac{1}{4} \int_2^3 \left( e^{uv} \Big\vert_{v=0}^{v=1}  \right) du \\
  &= \frac{1}{4} \int_2^3 \left( e^{u}-1  \right) du \\
  &= \frac{1}{4} \left( e^{u}-u  \right) \Big\vert_{u=2}^{u=3}  \\
  &= \frac{1}{4} \left( e^{3}-3 -e^2+2 \right) \\
  &= \frac{1}{4} \left( e^{3}-e^2-1 \right) \\
\end{align*}$$
Is this correct? 
I have some problem to affirm this because if i don't consider the condition $x \geq 0 $ in the domain $D$ the integral must be zero because the integrand function is odd respect the $x-$variable and the domain $D$ is symmetric respect the $y-$axis
 [/INST] You considered condition $x\geq 0$ implicitly when you canceled out $x$ and $\frac{1}{|4x|}$. Also condition $x\geq 0$ is neccesasry because it guarantees that $\Phi$ is bijective. As  for computations - everything is correct.
</s>