Non-inverting: zero at -5, pole at -3, HF gain 2

Required function: zero at $s = -5$, pole at $s = -3$, high-frequency gain 2:

$$T(s) = 2\,\frac{s+5}{s+3}$$

HF gain = 2, DC gain = $2 \times 5/3 = 3.333$.

Non-inverting configuration: $Z_1$ from the inverting input to ground, $Z_2$ from output to inverting input:

$$T(s) = 1 + \frac{Z_2}{Z_1}$$

Since the zero is larger than the pole and the HF gain is above 1, use $Z_1 = R_1$ and $Z_2 = R_a + (R_b \parallel C)$:

$$\begin{aligned} T(s) &= 1 + \frac{1}{R_1}\left(R_a + \frac{R_b}{1+sR_bC}\right) \\ &= \frac{R_1 + R_a}{R_1}\cdot\frac{s + \frac{1}{R_bC}\left(1 + \frac{R_b}{R_1+R_a}\right)}{s + \frac{1}{R_bC}} \end{aligned}$$

         +--Ra--+--Rb--+
         |      |      |
         |      +--||--+
         |         C   |
   +-----+-|-\         |
   |       |  >--------+--- Vo
  R1 Vin---|+/
   |
  GND

Matching:

$$\begin{aligned} 1 + \frac{R_a}{R_1} &= 2 \Rightarrow R_a = R_1 \\ \frac{1}{R_bC} &= 3 \\ 3\left(1 + \frac{R_b}{2R_1}\right) &= 5 \Rightarrow R_b = \frac{4}{3}R_1 \end{aligned}$$

Normalized ($C = 1$ F): $R_b = 0.3333\ \Omega$, $R_1 = R_a = 0.25\ \Omega$.

Check: DC gain $= 1 + \dfrac{R_a + R_b}{R_1} = 1 + \dfrac{0.5833}{0.25} = 3.333 = 2\times\frac53$. Correct.

Practical values ($k_m = 10^5$): $R_1 = R_a = 25\ \text{k}\Omega$, $R_b = 33.3\ \text{k}\Omega$, $C = 10\ \mu$F.

Answer: $R_1 = R_a = 25\ \text{k}\Omega$, $R_b = 33.3\ \text{k}\Omega$, $C = 10\ \mu$F.