$\int \frac{d z}{s^{2} \sqrt{s^{2}-9}}$

Step 1 ：Given the integral: \(\int \frac{d z}{s^{2} \sqrt{s^{2}-9}}\)

Step 2 ：The integral of the function with respect to z is: \(\frac{z}{s^{2} \sqrt{s^{2}-9}} + C\)

Step 3 ：\(\boxed{\frac{z}{s^{2} \sqrt{s^{2}-9}} + C}\)