Computing $\int_0^\infty \frac{dx}{(x^2 + 1)^2} $ with the residue theorem I am solving this integral and am stuck on proving an inequality, I believe I have the rest worked out.
My work:
First observe that the integrand is even, hence $$\int_0^\infty \frac{dx}{(x^2+1)^2} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2}$$
We now consider the integral on the r.h.s. above and integrate over the upper half circle with radius $R$ and the segment $[-R, R]$ on the real axis. We let $R$ tend towards infinity. Denoting this region by $D$, and the upper half circle by $C$, by Cauchy's Residue Theorem we have,
$$\int_{D} \frac{dz}{(z^2+1)^2} = 2\pi i \sum \text{Res} = \int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}$$
I first calculate the residues of the integrand in $D$. Notice there is a pole of order 2 at $z = i$. Hence,
$$\text{Res}_{z = i} = \lim_{z \rightarrow i} \Bigl( \frac{(z-i)^2}{(z^2+1)^2} \Bigr)' = \frac{1}{4i}$$
Thus,
$$\frac{\pi}{2} = \int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}$$
Now I would like to show that the complex integral on the r.h.s. tends to 0 as $R$ tends to infinity:
$$\int_C \frac{dz}{(z^2+1)^2} \leq \Biggl \vert \int_C \frac{dz}{(z^2+1)^2} \Biggr \vert \leq \sup f(X) \cdot L = \sup f(X) \cdot \pi R$$
This is as far as I have gotten. I know I have to play with the triangle inequality somehow to bound the integrand from above and determine $\sup f(x)$, but I cannot figure out how to do this.
One thought I had was to use the reverse triangle inequality since:
$$|(z^2+1)^2| = |z^4 +2z^2 + 1| \geq \bigl| |z^4| - |2z|^2 - 1\ \bigr|$$
so that,
$$\frac{1}{(z^2+1)^2} \leq \frac{1}{\bigl| |z^4| - |2z|^2 - 1\ \bigr|} = \frac{1}{\bigl| R^4 - 2R^2 - 1\ \bigr|}$$
This will wrap things up as
$$\sup f(x) \cdot R = \frac{\pi R}{\bigl| R^4 - 2R^2 - 1\ \bigr|}$$
which will tend to 0 as $R$ tends to infinity.
Putting it all together and diving by two,
$$\int_0^\infty \frac{dx}{(x^2 + 1)^2} = \frac{\pi}{4}$$
This is the correct numerical answer, but I am not sure if my use of the reverse triangle inequality is correct, can anyone confirm? Thanks!

The idea of your proof is correct, but there are some inaccuracies. First, it should be
$$
\frac{\pi}{2} = \int_R^R \frac{dx}{(x^2+1)^2} + \int_C \frac{dz}{(z^2+1)^2}
$$
for $R > 1$. You cannot replace the $R$ by $\infty$ in the first integral because the second integral depends on $R$. I would denote the half-circle with $C_R$ to emphasize the dependency on the radius.
Second,
$$
|(z^2+1)^2| = |z^4 +2z^2 + 1| \geq \bigl| |z^4| - |2z|^2 - 1 \bigr|
$$
is wrong as can be seen by substituting $z=i$. But
$$
|z^2+1| \ge |z^2| - 1 = R^2 - 1 > 0  \\
\implies |(z^2+1)^2| = |z^2+1|^2 \geq (R^2-1)^2
$$
for $R > 1$, so that
$$
 \sup_{z \in C_R} |f(z)| \cdot L(C_R) \le \frac{\pi R}{R^4-2R^2+1}
$$
converges to zero for $R \to \infty$. (You wrote $\sup f(x) \cdot R = \ldots$, which is not correct because we have only an upper bound for the supremum, not the exact value.)
Then
$$
\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2} = \lim_{R \to \infty}\int_R^R \frac{dx}{(x^2+1)^2}
= \frac \pi 2 - \lim_{R \to \infty} \int_{C_R} \frac{dz}{(z^2+1)^2}
= \frac \pi 2 \, .
$$
