1. **State the problem:** We need to evaluate the definite integral $$\int_0^1 \sqrt{17 + 4z^2} \, dz.$$\n\n2. **Formula and approach:** This integral involves a square root of a quadratic expression. A common method is to use a hyperbolic substitution or recognize it as a standard integral of the form $$\int \sqrt{a^2 + u^2} \, du.$$\n\n3. **Rewrite the integral:** Let $$a = \sqrt{17}$$ and $$u = 2z,$$ so $$du = 2 dz \Rightarrow dz = \frac{du}{2}.$$ When $$z=0,$$ $$u=0,$$ and when $$z=1,$$ $$u=2.$$\n\nThe integral becomes $$\int_0^2 \sqrt{a^2 + u^2} \cdot \frac{1}{2} du = \frac{1}{2} \int_0^2 \sqrt{17 + u^2} \, du.$$\n\n4. **Use the standard integral formula:**\n$$\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2} \sqrt{x^2 + a^2} + \frac{a^2}{2} \ln\left|x + \sqrt{x^2 + a^2}\right| + C.$$\n\n5. **Apply the formula:**\n$$I = \frac{1}{2} \left[ \frac{u}{2} \sqrt{u^2 + 17} + \frac{17}{2} \ln\left|u + \sqrt{u^2 + 17}\right| \right]_0^2.$$\n\n6. **Evaluate at the bounds:**\nAt $$u=2$$:\n$$\frac{2}{2} \sqrt{4 + 17} + \frac{17}{2} \ln\left(2 + \sqrt{21}\right) = 1 \cdot \sqrt{21} + \frac{17}{2} \ln(2 + \sqrt{21}).$$\nAt $$u=0$$:\n$$0 + \frac{17}{2} \ln(0 + \sqrt{17}) = \frac{17}{2} \ln(\sqrt{17}) = \frac{17}{2} \cdot \frac{1}{2} \ln(17) = \frac{17}{4} \ln(17).$$\n\n7. **Subtract to find the definite integral:**\n$$I = \frac{1}{2} \left( \sqrt{21} + \frac{17}{2} \ln(2 + \sqrt{21}) - \frac{17}{4} \ln(17) \right).$$\n\n8. **Simplify:**\n$$I = \frac{\sqrt{21}}{2} + \frac{17}{4} \ln(2 + \sqrt{21}) - \frac{17}{8} \ln(17).$$\n\n**Final answer:**\n$$\boxed{\int_0^1 \sqrt{17 + 4z^2} \, dz = \frac{\sqrt{21}}{2} + \frac{17}{4} \ln(2 + \sqrt{21}) - \frac{17}{8} \ln(17)}.$$