Z-transform of ZOH with plant 1/(s(s+1))

The ZOH transfer function is $G_h(s) = \dfrac{1-e^{-Ts}}{s}$. The factor $e^{-Ts}$ is a delay of one sampling period, which becomes $z^{-1}$. So the transform is $(1-z^{-1})$ times the z-transform of the plant divided by $s$:

$$F(z) = \mathcal{Z}\left[\frac{1-e^{-Ts}}{s}\cdot\frac{1}{s(s+1)}\right] = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right]$$

Partial fractions

$$\frac{1}{s^2(s+1)} = \frac{1}{s^2} - \frac{1}{s} + \frac{1}{s+1}$$

z-transform of each term

$$\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{Tz}{(z-1)^2} - \frac{z}{z-1} + \frac{z}{z-e^{-T}}$$

Multiply by $(1-z^{-1}) = (z-1)/z$

$$\begin{aligned} F(z) &= \frac{T}{z-1} - 1 + \frac{z-1}{z-e^{-T}} \\ &= \frac{T(z-e^{-T}) - (z-1)(z-e^{-T}) + (z-1)^2}{(z-1)(z-e^{-T})} \\ &= \frac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})} \end{aligned}$$

With $T = 1$ s

$e^{-1} = 0.3679$, so $T - 1 + e^{-T} = 0.3679$ and $1 - e^{-T} - Te^{-T} = 1 - 2(0.3679) = 0.2642$.

$$F(z) = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} = \frac{0.3679z^{-1} + 0.2642z^{-2}}{1 - 1.3679z^{-1} + 0.3679z^{-2}}$$

Answer: $F(z) = \dfrac{(T-1+e^{-T})z + (1-e^{-T}-Te^{-T})}{(z-1)(z-e^{-T})}$, which for $T = 1$ s is $\dfrac{0.3679z + 0.2642}{(z-1)(z-0.3679)}$.