I want to know what you think about the following solution to this problem: calculate, using the residue theory, the integral $$ \int_0^\infty \dfrac{1}{(x^2+9)^2}\ dx\ . $$
I took a path which is the union of the interval $[-R,R]$ of the real line and the semicircunference $C_R$ of radius $R$ in the semiplane with positive imaginary part. With these definitions, we can say that $$ \int_{-R}^{R} \dfrac{1}{(x^2+9)^2}\ dx + \int_{C_R} \dfrac{1}{(z^2+9)^2}\ dz = 2\pi i \sum_{k} \underset{z=z_k}{\operatorname{Res}} \dfrac{1}{(z^2+9)^2}\ , $$ where $z_k$ are the points where the function $f(z) = 1/(z^2+9)^2$ has an isolated singularity inside the region delimited by $C_R$ and the real interval $[-R,R]$.
In the case that $R>3$, the unique singularity inside that region is at $z= 3i$. Note that we can express $f(z)$ as $$ f(z) = \dfrac{\phi(z)}{(z-3i)^2}\ ,\text{where } \phi(z) = \dfrac{1}{(z+3i)^2}\ , $$ and the residue is therefore $$ \underset{z=3i}{\operatorname{Res}} f(z) = \left.\dfrac{d\phi}{dz} \right|_{z=3i} = \dfrac{-i}{108}\ . $$
After this, we can say that $$ \int_{-R}^{R}f(x)\ dx = \dfrac{\pi}{54} - \int_{C_R} f(z)\ dz \ . $$
Now, realize that $$ \dfrac{1}{|(z^2+9)^2|} \leq \dfrac{1}{|z^2+9|} \leq \dfrac{1}{|z|^2-9} = \dfrac{1}{R^2-9}\ ,\quad R>3\ , $$ which implies that $$ \left| \int_{C_R} f(z)\ dz \right| \leq \int_{C_R} \dfrac{1}{R^2-9}\ dz = \dfrac{\pi R}{R^2 - 9}\ , $$ and, finally $$ \lim_{R\to\infty} \dfrac{\pi R}{R^2 - 9} = 0 \Rightarrow \lim_{R\to\infty} \int_{C_R} f(z)\ dz = 0\ . $$
We conclude that $$ \int_{-\infty}^{\infty} \dfrac{1}{(x^2+9)^2}\ dx = \dfrac{\pi}{79} \Rightarrow \int_{0}^{\infty} \dfrac{1}{(x^2+9)^2}\ dx = \dfrac{\pi}{108} $$
Is this procedure correct or did I make any mistake along the process? I would appreciate any opinion.