# 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

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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.