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