how to evaluate $$\iint_{y\ge x^2+1}{dx\,dy\over{x^4+y^2}}$$
My solution: the initial intergral
$$ =2\int_0^\infty \left(\int_{x^2+1}^\infty {dy\over {x^4+y^2}}\right)\,dx = \int_0^\infty \int_{x^2+1}^\infty {{1\over{x^2} }d({y\over {x^2}})\over{1+({y\over x^2}})^2} $$
$$ = \int_0^\infty {1\over x^2} \left({\pi\over 2} - \arctan\left(1+{1\over x^2}\right) \right)dx, $$
I feel confused here. Can somebody give me other methods to solve the question?
You used a wrong way to write your double integral. You have to change the domain. Let $u=x,v=y-x^2$. Then your domain $D=\{(x,y): y\ge x^2+1\}$ can be changed into $D'=\{(u,v): -\infty< u<\infty, 1\le v<\infty\}$. Note $$\det\frac{\partial(u,v)}{\partial(x,y)}=1$$ and hence \begin{eqnarray*} \iint_{D}{dx\,dy\over{x^4+y^2}}&=&\iint_{D'}\frac{1}{u^4+(u^2+v)^2}dudv\\ &=&2\int_0^\infty\int_1^\infty\frac{1}{u^4+(u^2+v)^2}dvdu\\ &=&2\int_0^\infty\int_{u^2+1}^\infty\frac{1}{w^2+u^4}dwdu\\ &=&\int_0^\infty\frac{\pi-2\arctan(1+\frac{1}{u^2})}{u^2}du \end{eqnarray*} which can be evaluated by integration-by-parts formula as $\sqrt{2\sqrt{2}-2}\pi$.