First, we want a strategy for approximating \(\arctan 2\text{.}\) Our hints are that involves integrating \(\dfrac{1}{1+x^2}\text{,}\) which is the antiderivative of arctangent, and the number \(\dfrac{\pi}{4}\text{,}\) which is the same as \(\arctan(1)\text{.}\) With that in mind:

\begin{align*}
\int_1^2 \frac{1}{1+x^2} \dee{x}& = \arctan(2) - \arctan(1) = \arctan(2) - \frac{\pi}{4}\\
\text{So, }\qquad \arctan(2) &= \frac{\pi}{4} + \int_1^2 \frac{1}{1+x^2} \dee{x}\tag{$*$}\\
\end{align*}
We won't know the value of the integral exactly, but we'll have an approximation \(A\) bounded by some positive error bound \(\varepsilon\text{.}\) Then,

\begin{align*}
- \varepsilon &\leq \left(\int_1^2 \frac{1}{1+x^2} \dee{x} - A\right) \leq \varepsilon\\
A - \varepsilon &\leq \left(\int_1^2 \frac{1}{1+x^2} \dee{x}\right) \leq A + \varepsilon\\
\text{So, from ($*$), }\qquad \frac{\pi}{4}+A-\varepsilon& \leq \arctan(2) \leq \frac{\pi}{4}+A+\varepsilon
\end{align*}
Which approximation should we use? We're given the fourth derivative of \(\dfrac{1}{1+x^2}\text{,}\) which is the derivative we need for Simpson's rule. Simpson's rule is also usually quite efficient, and we're very interested in not adding up dozens of terms, so we choose Simpson's rule.

Now that we've chosen Simpson's rule, we should decide how many intervals to use. In order to bound our error, we need to find a bound for the fourth derivative. To that end, define \(N(x) = 24(5x^4-10x^2+1)\text{.}\) Then \(N'(x) = 24(20x^3-20x)= 480x(x^2-1)\text{,}\) which is positive over the interval \([1,2]\text{.}\) So, \(N(x) \leq N(2)=24(5\cdot 2^4 - 10\cdot 2^2+1)=984\) when \(1 \leq x \leq 2\text{.}\) Furthermore, let \(D(x)=(x^2+1)^5\text{.}\) If \(1 \leq x \leq 2\text{,}\) then \(D(x) \geq 2^5\text{.}\) Now we can find a reasonable value of \(L\text{:}\)

\begin{align*}
|f^{(4)}(x)|&=\left|\frac{25(5x^4-10x^2+1)}{(x^2+1)^5}\right| =
\left|\frac{N(x)}{D(x)}\right| \leq \frac{984}{2^5} = \frac{123}{4} = 30.75
\end{align*}
So, we take \(L=30.75\text{.}\)

We want \(\left[\dfrac{\pi}{4}+A-\varepsilon ,\, \dfrac{\pi}{4}+A+\varepsilon \right]\) to look something like \(\left[\dfrac{\pi}{4}+0.321,\, \dfrac{\pi}{4}+0.323\right]\text{.}\) Note \(\varepsilon\) is half the length of the first interval. Half the length of the second interval is \(0.001 = \frac{1}{1000}\text{.}\) So, we want a value of \(\varepsilon\) that is no larger than this. Now we can find our \(n\text{:}\)

\begin{align*}
\dfrac{L(b-a)^5}{180\cdot n^4} &\leq \frac{1}{1000}\\
\dfrac{30.75}{180\cdot n^4} &\leq \frac{1}{1000}\\
n^4 & \geq \frac{30.75\times 1000}{180}\\
n & \geq \sqrt[4]{\frac{30750}{180}}\approx 3.62
\end{align*}
So, we choose \(n=4\)), and are guaranteed that the absolute error in our approximation will be no more than \(\dfrac{30.75}{180\cdot 4^4} \lt 0.00067\text{.}\)

Since \(n=4\text{,}\) then \(\De x = \dfrac{b-a}{n}=\dfrac{1}{4}\text{,}\) so:

\begin{equation*}
x_0 = 1 \qquad x_1=\frac{5}{4} \qquad x_2 = \frac{3}{2} \qquad x_3 = \frac{7}{4} \qquad x_4=2
\end{equation*}
Now we can find our Simpson's rule approximation \(A\text{:}\)

\begin{align*}
\int_0^1 \frac{1}{1+x^2} \dee{x}&\approx \frac{\De x}{3}\big[
f(x_0) + 4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)
\big]\\
&=\frac{1/4}{3}\big[
f(1) + 4f(5/4)+2f(3/2)+4f(7/4)+f(2)
\big]\\
&=\frac{1}{12}\left[
\frac{1}{1+1} + \frac{4}{25/16+1}+\frac{2}{9/4+1}+\frac{4}{49/16+1}+\frac{1}{4+1}
\right]\\
&=\frac{1}{12}\left[
\frac{1}{2} + \frac{4\cdot 16}{25+16}+\frac{2\cdot 4}{9+4}+\frac{4\cdot 16}{49+16}+\frac{1}{5}
\right]\\
&=\frac{1}{12}\left[
\frac{1}{2} + \frac{64}{41}+\frac{8}{13}+\frac{64}{65}+\frac{1}{5}
\right]\\
&\approx 0.321748=A
\end{align*}
As we saw before, the error associated with this approximation is at most \(\dfrac{30.75}{180\cdot 4^4} \lt 0.00067=\varepsilon\text{.}\) So,

\begin{align*}
&A-\varepsilon \quad&&\leq\quad && \int_1^2\frac{1}{1+x^2} \dee{x} &\leq \quad &A+\varepsilon&\\
\Rightarrow\qquad&0.321748 - 0.00067 \quad&&\leq\quad && \int_1^2\frac{1}{1+x^2} \dee{x} \quad &\leq \quad &0.321748 + 0.00067&\\
\Rightarrow\qquad&0.321078 &&\leq\quad & &\int_1^2\frac{1}{1+x^2} \dee{x}\quad &\leq \quad &0.322418&\\
\Rightarrow\qquad&0.321 &&\leq\quad &&\int_1^2\frac{1}{1+x^2} \dee{x}\quad &\leq \quad &0.323&\\
\Rightarrow\qquad&\frac{\pi}{4}+0.321 && \leq\quad &&\int_1^2\frac{1}{1+x^2} \dee{x}+\frac{\pi}{4}\quad &\leq \quad &\frac{\pi}{4}+0.323&\\
\Rightarrow\qquad&\frac{\pi}{4}+0.321 && \leq\quad &&\arctan(2) &\leq \quad &\frac{\pi}{4}+0.323&
\end{align*}
This is precisely what we wanted to show.