Is the following integral positive? For each positive integer $n$, consider the following intgral:
$$\int_0^\infty\int_0^\infty\frac{(x^4-y^2)x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy.$$
I want to know if there is any easy way to see that it's positive. If there is no easy proof to show that it's positive, I am satisfied if I know that it is positive. I don't need to know the exact value of the integral. I try to write the integral as
$$\int_0^\infty\int_0^\infty\frac{x^{2n+1}}{[(1+x^2)^2+y^2]^{n+1}}dxdy-
\int_0^\infty\int_0^\infty\frac{2y^2x^{2n+1}}{(x^4+y^2)[(1+x^2)^2+y^2]^{n+1}}dxdy,$$
but it seems to me that it doesn't help much. 
 A: Start with the substitution suggested by @alex.jordan: $u=\sqrt{x}, v=y$. Then think of $(u,v)$ as Cartesian coordinates with the integral over the first quadrant. Change to polar coordinates $u=r\cos\theta, v=r\sin\theta$ to obtain
$$ \hbox{your integral } = I_n = \frac12\int_0^\infty r^{n+1} dr \int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta.$$
Split 
$$J_n=\int_0^{\pi/2} \frac{\cos^n\theta(\cos^2\theta-\sin^2\theta)}{(r^2+2r\cos\theta+1)^{n+1}}d\theta$$
at $\theta=\pi/4$ to obtain integrands that are obviously positive and negative. In the subinterval $\theta\in[\pi/4,\pi/2]$ change the variable to $\theta^\prime=\pi/2-\theta$ - the domain for the second integral is now $\theta^\prime\in[0,\pi/4]$ - and then change the name of $\theta^\prime$ to $\theta$. Recombine the two integrals to obtain
$$ J_n=\int_0^{\pi/4} (h_n(r,\cos\theta)-h_n(r,\sin\theta))(\cos^2\theta-\sin^2\theta)d\theta$$
where 
$$h_n(r,z)=\frac{z^n}{(r^2+2rz+1)^{n+1}}.$$
For $n\geq 2$, this is an increasing function of $z$ on $[0,1]$ which is sufficient to show that the integrand in the second representation of $J_n$ is positive. I think this deals with everything but the case $n=1$. 
A: It's an idea only, but too long in the comment (I run out of the space.)(Sorry.)
Following the idea of @alex.jordan, introduce the new variables $X:=\sqrt{x}, Y:=\sqrt{y}$ by steps. Then we obtain
$$
\int_0^{\infty}\int_0^{\infty}\frac{(X^2-Y)X^n}{4(X^2+Y)((1+X)^2+Y)^{n+1}\sqrt{Y}}\,dXdY
$$
Now we use the transformation $X^2+Y=u,(1+X)^2+Y=v$.
Then the abs'value of Jacobi-det is $\frac{1}{2}$ and we get
$$
\int_0^{\infty}\int_1^{\infty} \frac{\left(\frac{1}{2}-v+\frac{1}{2}u^2-uv+\frac{1}{2}v^2\right)\left(-\frac{1}{2}-\frac{1}{2}u+\frac{1}{2}v\right)^n}{8uv^{n+1}\sqrt{-\frac{1}{4}+\frac{1}{2}u+\frac{1}{2}v-\frac{1}{4}u^2+\frac{1}{2}uv-\frac{1}{4}v^2}}\,dvdu.
$$
The advantage of the last integral, that the terms containing $n$ in the exponent are simple. But I have to think further ...
