= Solution
By the <Bromwich inversion formula>, for $t>0$ and $\gamma>0$,
$$
f(t)=\frac1{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s^2(s+1)^2}\,ds.
$$
Close the upward vertical segment by a left semicircle of radius $R$ centred at $\gamma$. On that arc $\operatorname{Re}s\le\gamma$, so $|e^{st}|\le e^{\gamma t}$, while $|s^2(s+1)^2|$ is bounded below by a constant times $R^4$ for large $R$. Its length is $\pi R$, hence the arc <integral> tends to zero as $O(R^{-3})$. The <contour> is positively oriented and encloses the <double poles> at $0$ and $-1$.
Their <residues> are
$$
\operatorname{Res}_{s=0}\frac{e^{st}}{s^2(s+1)^2}=\left.\frac d{ds}\frac{e^{st}}{(s+1)^2}\right|_{s=0}=t-2,
$$
and
$$
\operatorname{Res}_{s=-1}\frac{e^{st}}{s^2(s+1)^2}=\left.\frac d{ds}\frac{e^{st}}{s^2}\right|_{s=-1}=(t+2)e^{-t}.
$$
The <residue theorem> therefore yields
$$
\boxed{f(t)=t-2+(t+2)e^{-t}\qquad(t\ge0).}
$$
The value at zero follows by <continuity> and is $f(0)=0$.
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