How to design compensator for control system
As suggested by Chu, a simple extra zero should do the job. Why adding another pole at the origin to cancel the static error since the plant already has one? The zero will thus be positioned to select the right crossover \$f_c\$ and the desired phase margin.

The transfer function of the plant can be rewritten in a low-entropy format, showing the time constants at work: \$H(s)=\frac{1}{\frac{s}{\omega_{po}}(1+\frac{s}{\omega_p})}\$ with \$\omega_{po}=445\,s^{-1}\$ and \$\omega_{p}=2.25\,ms^{-1}\$.



To counteract the pole position and cross the 0-dB axis with a -1-slope, we can position the zero for a 60° phase margin (PM). The zero position is obtained via the following expression: \$f_z=\frac{f_c}{tan^{-1}(PM-arg(H(f_c)-180°)}\$.

The gain of the compensator is calculated to compensate the pole attenuation at the selected crossover \$f_c\$: \$G_0=\frac{1}{|H(f_c)|}\frac{1}{\sqrt{1+(\frac{f_c}{f_z})^2}}\$. A Mathcad sheet tells us if we got it right:



One we have these data on hand, we can test the closed-loop response to a 1-V step;



The response is perfect and the error is null as expected.