Continuity of $\frac{1}{x^2}$ given the metric $d(a,b)= \left| \int_a^b \frac{1}{x^2 +1} dx\right| $ on the extended real line. Let $y = \frac{1}{x^2}$ given the metric $d(a,b)= \left| \int_a^b \frac{1}{x^2 +1} dx\right|  $ on a part of the extended real line $[0, +\infty]$ where $$ f(0) = \infty, \; \; f(x) = \frac{1}{x^2}, \; \; f(\infty)=0 $$
So I start with the $\epsilon - \delta$ proof format. The integral metric given can be written as: $ d(a,b) = |\arctan(b) - \arctan(a)|$. Given this, we want to say that knowing: $$ |\arctan(b) - \arctan(a)| < \delta $$
we must show that for all $\epsilon > 0$, $$ \left|\arctan(\frac{1}{b^2}) - \arctan(\frac{1}{a^2})  \right|<\epsilon $$
This is how far I got: $$ \left|\arctan(\frac{1}{b^2}) - \arctan(\frac{1}{a^2})  \right| = \left|-\arctan(b^2) + \arctan(a^2)\right| = \\ |-1|\left|\arctan(b^2) - \arctan(a^2)\right| = \left|\arctan(b^2) - \arctan(a^2)\right|  $$
But I don't really know how to proceed from there and extract $ |\arctan(b) - \arctan(a)|$ out of it so that I can relate $\epsilon$ and $\delta$.
Ideas?
 A: In fact $f(x):=1/x^2$ is uniformly continuous relative to the metric $d$. To show this, we first establish inequalities showing that the metric $d(a,b)$ is comparable to the standard metric $|a-b|$:
For all $x\ge0$ we have
$$\frac1{(x+1)^2}\le\frac1{x^2+1}\le1.$$
Integrating these from $a$ to $b$, conclude that
$$\frac{|a-b|}{(a+1)(b+1)}\stackrel{(1)}\le d(a,b)\stackrel{(2)}\le|a-b|.$$
To prove uniform continuity we want to assert $d(f(a),f(b))\le K d(a,b)$, where $K$ is a constant independent of $a$ and $b$.
Assume $a<b$. There are two overlapping cases to consider: $a<b\le m$ and $M\le a<b$, where $M<m$ (sic) are constants with $m$ chosen big enough and $M$ chosen small enough to cover all possibilities. (This is doable because if $d(a,b)<\delta$, then there is a limit to how far $b$ is from $a$.)
For the "small" case $a<b\le m$, we can argue
$$
\begin{align}
d(f(a),f(b))=d(1/a^2,1/b^2)&=d(a^2,b^2)\\
&\stackrel{(2)}
\le|a^2-b^2|\\
&=(a+b)|a-b|\\
&\stackrel{(1)}\le(a+b)(a+1)(b+1)d(a,b)\\
&\le(2m)(m+1)^2d(a,b).
\end{align}$$
Note the above argument is valid for the case $a=0$, where we are attempting to bound $d(\infty,1/b^2)$.
The "large" case $M\le a<b$ can be reduced to the small case by reciprocation:
$
d(1/a^2,1/b^2)=d(a_1^2,b_1^2)$ where $a_1:=1/b$ and $b_1:=1/a$ are such that $a_1<b_1\le 1/M$.
