Does this integral converge $\int_{\mathbb{R}^2}\frac{1}{x^4y^4+1}\ dxdy $? I want to find out whether this integral is convergent or not.
$$\int_{\mathbb{R}^2}\frac{1}{x^4y^4+1}\ dxdy $$
I've tried to calculate it using the following variable changement, but it does'nt work i guess.$(x,y)=(r\cdot \cos(\theta),r\cdot \sin(\theta))$.
I also though of comparing the general term to another one that converge but i couldn't find. 
 A: 
The integral diverges.

To see this, note that the integrand is positive.  Therefore, the value of the integral over $\mathbb{R}^2$ is an upper bound of the value of the integral over any subset of $\mathbb{R}^2$.
Let $S$ be the set given by $S=\{(x,y)|xy\le 1, x>0, y>0\}$.  Then, we have
$$\begin{align}
\iint_{\mathbb{R}^2}\frac{1}{1+x^4y^4}\,dx\,dy&\ge \iint_{S}\frac{1}{1+x^4y^4}\,dx\,dy\\\\
&\ge\frac12\iint_{S}\,dx\,dy
\end{align}$$
But the area of $S$ is clearly infinite.

Therefore, we conclude the integral of interest diverges.

A: Another way to show divergence: The substitution $x=u/|y|$ gives
$$\int_{-\infty}^\infty\int_{-\infty}^\infty{1\over x^4y^4+1}dxdy=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty{1\over u^4+1}\cdot{du\over |y|} \right)dy=\left(\int_{-\infty}^\infty{du\over u^4+1}\right)\left(\int_{-\infty}^\infty{dy\over|y|}\right)$$
The improper integral over $u$ converges (to a nonzero value), but the one over $y$ diverges.
Remark: The use of $|y|$ instead of just $y$ in the substitution is to keep the limits of integeration running from $-\infty$ to $\infty$ for all $y$.  Technically the substitution is invalid for $y=0$, but that's only a single point in the integral over $y$.
My thanks to user Simply Beautiful Art for pointing out an error in the original version of the answer.
A: This answer is wrong, as the integral diverges.
Since I missed the case of $|x|>1$ and $|y|<1$ and vice versa.

Hint:
$$\int\frac1{x^4y^4+1}\ dx\ dy<\int\frac1{x^4y^4}\ dx\ dy$$
and use it to show convergence for when $|x|>1$ and $|y|>1$.
For $|\cdot|\le1$, show that it is finite.
