Lorentz invariance of the action for free relativistic particle I tried to check the Lorentz invariance of the standard special relativity action for free particle directly: ($c=1$)
$$
S=\int L dt=-m\int\sqrt{1-v^{2}}dt 
$$
Lorentz boost:
$$ dt=\frac{dt^{'}+udx^{'}}{\sqrt{1-u^{2}}}$$,  $$dx=\frac{udt^{'}+dx^{'}}{\sqrt{1-u^{2}}}$$,  $$v=\frac{dx}{dt}=\frac{u+v^{'}}{1+uv^{'}}$$
Substitute these expressions into the action:
$$
S^{'}=-m\int \sqrt{1-\left(\frac{u+v^{'}}{1+uv^{'}} \right)^{2}} dt=-\int \sqrt{1-\left(\frac{u+v^{'}}{1+uv^{'}} \right)^{2}} \left(\frac{dt^{'}+udx^{'}}{\sqrt{1-u^{2}}}\right)
$$
$$
S^{'}=-m\int \left(\frac{\sqrt{1-\left(\frac{u+v^{'}}{1+uv^{'}} \right)^{2}}}{\sqrt{1-u^{2}}} \right)dt^{'}-m\int \left(\frac{u\sqrt{1-\left(\frac{u+v^{'}}{1+uv^{'}} \right)^{2}}}{\sqrt{1-u^{2}}}\right)dx^{'} 
$$
But I know that action of this type is invariant under lorentz transformations, so we want to get that
$$
S^{'}=-m\int \sqrt{(1-v'^{2})} dt^{'}
$$
But it doesnt seem to me that these expressions for action are equal. So I'm confused.
Maybe there is a mistake in my calculations?
 A: joshphysics's answer was excellent, and I'd like to draw attention to another aspect of the question.
Your action,
$S=-m\int\sqrt{1-u^{2}}dt$
Can be written:
$S=-m\int\sqrt{(dt)^{2}-(udt)^{2}}$
$=-m\int\sqrt{(dt)^{2}-(dx)^{2}} = -m\int ds$
And the transformed action $S' = -m \int ds'$.
So, for the action to be Lorentz invariant, it suffices to show that the interval, $s$, is Lorentz invariant, i.e. $s=s'$:
Say $s^2 = t^2 - x^2$; $x' = \gamma (x-vt)$; $t' = \gamma (t-vx)$; and finally $\gamma = (1-v^2)^{-1/2}$. Then,
$(s')^2 = (t')^2 - (x')^2$
$=\gamma^2((t-vx)^2 - (x-vt)^2) = \gamma^2(t^2 +v^2x^2 - x^2+ -v^2t^2)$
(Cross terms cancel)
$= (1-v^2)^{-1}(t^2(1-v^2) -x^2(1-v^2)) = t^2-x^2$
So $ds'=ds$, and consequently your action is Lorentz invariant.
A: I too will set $c=1$ and I'll ignore the mass and minus sign outside of the integral for simplicity since they don't affect Lorentz-invariance.  
Let $x^\mu(\lambda) = (t(\lambda), x(\lambda)$) denote a parameterized path in two-dimensional Minkowski space.  We can mathematically define the action of Lorentz transformations on such parameterized paths by
\begin{align}
  t'(\lambda) &= \gamma_u(t(\lambda)-ux(\lambda)) \\
  x'(\lambda) &= \gamma_u(x(\lambda) - ut(\lambda)) 
\end{align}
Physically, the transformed spacetime path defined in this way corresponds to what an observer in another inertial frame (with the inertial frame moving at velocity $u$) would measure.  The claim of Lorentz invariance of the free particle is simply that for a given parameterized path, if we define the following: action functional
$$
  S[t, x] = \int d\lambda \sqrt{\frac{dt}{d\lambda}^2 - \frac{dx}{d\lambda}^2}
$$
then
$$
  S[t', x'] = S[t, x]
$$
to see this, we simply compute
\begin{align}
  S[t', x'] 
&= \int d\lambda \sqrt{\frac{dt'}{d\lambda}^2 - \frac{dx'}{d\lambda}^2} \\
&= \int d\lambda \sqrt{\gamma_u^2\left(\frac{dt}{d\lambda} - u\frac{dx}{d\lambda}\right)^2 - \gamma_u^2\left(\frac{dx}{d\lambda} - u\frac{dt}{d\lambda}\right)^2} \\
&= \int d\lambda \sqrt{\gamma_u^2(1-v^2)\left(\frac{dt}{d\lambda}^2 -\frac{dx}{d\lambda}^2\right)} \\
&= \int d\lambda \sqrt{\frac{dt}{d\lambda}^2 -\frac{dx}{d\lambda}^2} \\
&= S[t,x]
\end{align}
Making contact with your notation:
Note that if $t(\lambda) = \lambda$, then $\lambda$ corresponds to the time as measured in the unprimed inertial frame.  In this case we could denote $v = \frac{dx}{dt}= \frac{dx}{d\lambda}$ and note that $\frac{dt}{d\lambda}= 1$ so that
$$
  S = \int dt\sqrt{1-v(t)^2}
$$
In this case, we would have additionally that
\begin{align}
  t'(t) &= \gamma_u(t- u x(t))\\
  x'(t) &= \gamma_u(x(t) - u t)
\end{align}
and you can go through the same analysis with the action.
