First integral of relativistic Euler-Lagrange equations Connsider a pseudo-Riemannian ($4$-dimensional) manifold $M$ with a pseudometric $g_{ab}$. The Lagrangian of a free particle in $M$ (in analogy to the flat case) is
$$\mathcal L=\frac{1}{2}g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}$$
where $\tau$ is the proper time and $x^a$ are the coordinates respect the charts. 
Now on my relativity book I read:

The four-velocity identity $g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}=-c^2$ is a first integral of the relativistic Euler-Lagrange equations for free material particles.

Now what does this phrase mean? What is a first integral for the Euler-Lagrange equations? Moreover why $g_{ab}\frac{dx^a}{d\tau}\frac{dx^b}{d\tau}=-c^2$?
Thanks in advance
 A: I) A first integral to a system of second-order ODEs 
$$ \tag{1} f^i(\ddot{q},\dot{q},q,t)~=~0$$ 
is usually an equation of the form
$$\tag{2} g(\dot{q},q,t)~=~0,$$ 
that (when differentiated wrt. the parameter $t$) produces an equation 
$$\tag{3} \left(\frac{\partial }{\partial t}
+\dot{q}^i\frac{\partial }{\partial q^i}
+\ddot{q}^i\frac{\partial }{\partial \dot{q}^i}\right)g(\dot{q},q,t)~=~0,$$ 
that in turn holds just as a consequence of eq. (1). 
Eqs. (1) and (2) are typically the equations of motion and an energy conservation statement, respectively. 
II) Consider next a massive relativistic point particle in a curved spacetime $(M,g)$ with Lagrangian density 
$$ \tag{4}{\cal L}~=~ g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}$$ 
[in signature $(+,-,-,-)$]. The first integral (of the corresponding Euler-Lagrange equations=geodesic equations) is that ${\cal L}$ is a constant of motion 
$$ \tag{5} {\cal L}~=~ \text{const}.$$
The equation mentioned by OP
$$ \tag{6} {\cal L}~=~ c^2$$
means that the length of the $4$-velocity is $c$. (That this is true can e.g. be seen by choosing a rest frame of the massive particle.) This in turn implies that the travelled arclength $\Delta s$ in the spacetime $(M,g)$ is $c\Delta\tau$, where $\tau$ is the proper time. 
The transition from eq. (5) to eq. (6) can be viewed as a scaling of the free parameter in (4), so that it becomes the proper time $\tau$.
A: I think it's misleading for the text to refer to the four-velocity identity as just another first integral of the equations of motion.  Instead, I'll argue that, in this context, this identity is better thought of as an initial condition.
The same issue arises in special relativity, so I'll first stay in flat-land.  There the Euler-Lagrange equations of motion are that the free particle's four-velocity $u^\mu$ is constant:
$$ \frac{d u^\mu}{d \tau} = 0 $$
but they do not fix its magnitude $ \eta_{\mu \nu} u^\mu u^\nu $.  If this quantity were a first integral like energy, it could take on arbitrary values.  But it's not, and it doesn't.
As Qmechanic describes, the four-velocity identity states that the squared magnitude of the four-velocity is a fundamental invariant of a material particle's motion in special relativity, true in all Lorentz frames and at all times:
$$ \eta_{\mu \nu} u^\mu u^\nu =\left| {\frac{dx}{d \tau}} \right|^2 - c^2 \left( \frac{dt}{d \tau} \right)^2 = \gamma^2 v^2 - \gamma^2 c^2 = - c^2 $$
The initial conditions for the equations of motion must be selected to comply with this identity.
Goldstein has some discussion of the subtleties involved in formulating covariant Lagrangians in special relativity.

Edit:
Generalizing to general relativity, Weinberg (Gravitation and Cosmology, Chapter 3.3 and 7.1) shows that, at any space-time point $P$, for arbitrary coordinates $x^\mu$ with metric $g$, the four-velocity is constant:
$$ \frac{d}{d\tau} \left\{ g_{\mu \nu} \frac{dx^\mu}{d \tau} \frac{dx^\nu}{d \tau}\right\} = 0  $$
if there exists a local coordinate system $y^\mu$ whose metric $h$ satisfies these conditions:
$$ h_{\mu \nu}(P) = \eta_{\mu \nu}  $$ 
$$ \left( \frac{\partial h_{\mu \nu}(y)}{\partial y^\gamma} \right)_{y=P} = 0 $$
It's not enough that the local coordinates' metric is "flat" at $P$; it must be stationary there as well.  Then, with the initial conditions setting the expression to $-c^2$, it will remain at that value, and the four-velocity identity will be satisfied.
