Why does $\frac{x^{2}}{1+x^{2}}$ diverge? 
Why does $\frac{x^{2}}{1+x^{2}}$ diverge?

I am trying to show that the integral $2\int\int_{\mathbb{R}^{+}}\frac{x^{2}}{(1+x^{2})(1+y^{2})}dxdy$ diverges. The solution says it diverges because $\frac{x^{2}}{1+x^{2}}\geq \frac{1}{2}$. What does this mean? What theorem is being used in analysis?
 A: It's not the function $f(x)=\frac{x^2}{1+x^2}$ that diverges (at $\infty$): indeed its limit at $\infty$ is $1$.
It's the integral
$$
\int_0^\infty \frac{x^2}{1+x^2}\,dx
$$
that diverges, for the very reason that the limit of the function at $\infty$ is not zero.
Since the limit is $1$, there exists $k>0$ such that, for $x>k$, $f(x)>1/2$. Hence, for $t>k$,
$$
\int_0^{t}f(x)\,dx=\int_0^k f(x)\,dx+\int_k^t\frac{1}{2}\,dx=
\int_0^k f(x)\,dx+\frac{1}{2}(t-k)
$$
and the last term of course diverges for $t\to\infty$.
A: The theorem being used is that if $f(x,y) \geq g(x,y)$, and if $\iint_R g(x,y)\,dx\,dy$ diverges, then $\iint_R f(x,y)\,dx\,dy$ diverges, too.
In your case, since
$$\frac{x}{1+x^2} \geq \frac{1}{2}$$
it follows that your integrand satisfies
$$f(x,y) = \frac{x^2}{(1+x^2)(1+y^2)} \geq \frac{1}{2}\frac{1}{1+y^2} = g(x,y)$$
Since the integral
$$\int_0^\infty \int_0^\infty \frac{1}{2}\frac{1}{1+y^2}\,dx\,dy$$
diverges (check this yourself), it follows that your original integral diverges.
A: $\dfrac{x^2}{x^2+1} =1-\dfrac{1}{x^2+1};$
For $x \ge 1$:  
$x^2+1 \ge 2$, and
$f(x):= \dfrac{x^2}{x^2+1} \ge 1/2.$
$\displaystyle \int_{1}^{t} (1/2)dx  \le \int_{1}^{t}f(x)dx$;.
$(1/2)(t-1) \le \displaystyle \int_{1}^{t}f(x)dx$, 
divergent for $t \rightarrow \infty.$
