1) A constant of motion
$f(z,t)$ is a (globally defined, smooth) function $f:M\times [t_i,t_f] \to
\mathbb{R}$ of the dynamical variables $z\in M$ and time $t\in[t_i,t_f]$,
such that the map $$[t_i,t_f]~\ni ~t~~\mapsto~~f(\gamma(t),t)~\in~ \mathbb{R}$$
doesn't depend on time for every solution curve $z=\gamma(t)$ to the equations of motion of the system.
An integral of motion/first integral
is a constant of motion $f(z)$ that doesn't depend explicitly on time.
2) In the following let us for simplicity restrict to the case where the system is a finite-dimensional autonomous$^1$ Hamiltonian system with Hamiltonian $H:M \to \mathbb{R}$ on a $2N$-dimensional symplectic manifold $(M,\omega)$.
Such system is called (Liouville/completely) integrable if
there exist $N$ functionally independent$^2$, Poisson-commuting, globally defined functions $I_1, \ldots, I_N: M\to \mathbb{R}$, so that the Hamiltonian $H$ is a function of $I_1, \ldots, I_N$, only.
Such integrable system is called maximally superintegrable if
there additionally exist $N-1$ globally defined integrals of motion $I_{N+1}, \ldots, I_{2N-1}: M\to \mathbb{R}$, so that the combined set $(I_{1}, \ldots, I_{2N-1})$ is functionally independent.
It follows from Caratheodory-Jacobi-Lie theorem that every finite-dimensional autonomous Hamiltonian system on a symplectic manifold $(M,\omega)$ is locally maximally superintegrable in sufficiently small local neighborhoods around any point of $M$ (apart from critical points of the Hamiltonian).
The main point is that (global) integrability is rare, while local integrability is generic.
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$^1$ An autonomous Hamiltonian system means that neither the Hamiltonian $H$ nor the symplectic two-form $\omega$ depend explicitly on time $t$.
$^2$ Outside differential geometry $N$ functions $I_1, \ldots, I_N$ are called functionally independent if $$\forall F:~~ \left[z\mapsto F(I_1(z), \ldots, I_N(z)) \text{ is the zero-function} \right]~~\Rightarrow~~ F \text{ is the zero-function}.$$
However within differential geometry, which is the conventional framework for dynamical systems, $N$ functions $I_1, \ldots, I_N$ are called functionally independent if $\mathrm{d}I_1\wedge \ldots\wedge \mathrm{d}I_N\neq 0$ is nowhere vanishing. Equivalently, the rectangular matrix $$\left(\frac{\partial I_k}{\partial z^K}\right)_{1\leq k\leq N, 1\leq K\leq 2N}$$ has maximal rank in all points $z$. If only $\mathrm{d}I_1\wedge \ldots\wedge \mathrm{d}I_N\neq 0$ holds a.e., then one should strictly speaking strip the symplectic manifold $M$ of these singular orbits.
Best Answer
I think the source of your confusion is mathematics, not physics. It is important here (and L&L did mention this) that the system of differential equations is autonomous. If this is the case, than along with solution $q_i=q_i(t,C_1,\dots,C_{2s}),$ it has a solution $q_i=q_i(t-t_0,C_1,\dots,C_{2s}).$ Because the former is the general solution, the later must reduced to it, that is, it should be $$q_i(t-t_0,C_1,\dots,C_{2s})=q_i(t,C_1'(C_1,\dots,C_{2s},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s},t_0)).$$ Putting $C_{2s}=0$, yields
$$q_i(t-t_0,C_1,\dots,C_{2s-1})=q_i(t,C_1'(C_1,\dots,C_{2s-1},t_0),\dots,C_{2s}'(C_1,\dots,C_{2s-1},t_0)).$$
For example, in the case of a free 1D motion $x=C_1t+C_2$. Making the shift in time one has: $$x=C_1t+(C_2-C_1t_0)$$ (expression in brackets is $C_2'$). And letting $C_2=0$ we find $$x=C_1(t-t_0).$$