complex analysis - 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?