Residue Calculus (Computing an Improper Integral) 
Use residue calculus to compute the integral $\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz$

My solution
If we add to the interval $I_{R}=[-R,R]$ add the semicircle $\gamma_{R}$ in the upper half plane with centre at the origin and radius $R$, then we obtain a closed contour $\Gamma_{R}$ (which we consider to be oriented in the positive direction over which the integral $f(z)=\frac{1}{(z^{2}+25)(z^{2}+16)}$ can be computed using the Cauchy's Residue theorem. 
\begin{align}
\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz&=\lim_{R \to \infty}\int_{\Gamma_{R}}f(z)dz \\ &=2\pi i \sum_{k=1}^{2}\text{Res}(z_{k}), \quad \Im(z_{k})>0
\end{align}
In this case, we have two simple poles at the points $z_{1}=4i$ and $z_{2}=5i$. The residues are computed as 
\begin{align}
\text{Res}(z_{1})&=\frac{1}{(z^{2}+25)D(z^{2}+16)}\big|_{z=4i} \\
&=\frac{1}{(z^{2}+25)2z}\big|_{z=4i} \\
&=\frac{1}{9\cdot8i} \\
&=\frac{1}{72i} \\
&=-\frac{i}{72}
\end{align}
Similarly, we get for $z_{2}$ that
\begin{align}
\text{Res}(z_{2})&=\frac{1}{D(z^{2}+25)(z^{2}+16)}\big|_{z=5i} \\
&=\frac{1}{2z(z^{2}+16)}\big|_{z=5i} \\
&=\frac{1}{10i \cdot (-9)} \\
&=-\frac{1}{90i} \\
&=\frac{i}{90}
\end{align}
Thus, we get
\begin{align}
\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz &=2\pi i \left(\text{Res}(z_{1})+\text{Res}(z_{2})\right) \\
&=2 \pi i \left(-\frac{i}{360}\right) \\
&=\frac{\pi}{180}
\end{align}
Are there other methods one could use to arrive at this answer?

Yes. Simply rewrite $$\int \frac{1}{(z^2+16)(z^2+25)} \mathrm{d}z= -\frac{1}{225} \int \frac{1}{\frac{z^2}{25}+1} \mathrm{d}z +\frac{1}{144} \int \frac{1}{\frac{z^2}{16}} \mathrm{d}z$$ and substitute $u=\frac{z}{5}$ for the first integral and $v=\frac{z}{4}$ for the second.
