<s>[INST] Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$ 
Integrate $\frac{x^3-3xy^2}{(x^2+y^2)^2}$ over $1\leq x^3-3xy^2\leq2,2\leq3x^2y-y^3\leq4$ and $x,y\geq0.$

Let $R$ be the region. I substitute $u=x^3-3xy^2,v=3x^2y-y^3$. $\det J(x,y)=9(x^2+y^2)^2.$ Also notice that $(x+yi)^3=u+vi$, so $(x^2+y^2)^3=u^2+v^2$.
Then the rquired integral
$$=\int_R\frac{x^3-3xy^2}{9(x^2+y^2)^4}9(x^2+y^2)^2dxdy$$
$$=\int^4_2\int^2_1\frac u{9(u^2+v^2)^{4/3}}dudv$$
$$=\int^4_2\frac 1 6[\frac 1 {(1+v^2)^{1/3}}-\frac 1{(4+v^2)^{1/3}}]dv$$
This has no closed form solution. I wonder where I did wrong... Thank you.
 [/INST] I think that your calculations are correct. The final result is given in terms of an hypergeometric function.
You may make use of 
$$
\int \frac{\mathrm{d} u}{(a^2+u^2)^{1/3}} = \frac{u}{a^{2/3}} \, {}_2F_1 \left( \frac{1}{3}, \frac{1}{2}; \frac{3}{2}; -\frac{u^2}{a^2} \right) \, , 
$$
to find a closed form of your integral.
Cheers,
R
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