Solving the soliton equation without energy In this passage from Srednicki's Quantum Field Theory (page 576)

The solution of interest is time independent, so we can set $\dot\varphi = 0$. We can also rewrite the remaining terms in $E$ as
  \begin{align}
E &= \int_{-\infty}^{+\infty}\mathrm{d}x\Bigl[\tfrac{1}{2}\Bigl(\varphi' - \sqrt{2V(\varphi)}\Bigr)^2 + \sqrt{2V(\varphi)}\varphi'\Bigr] \\
&= \int_{-\infty}^{+\infty}\mathrm{d}x \tfrac{1}{2}\Bigl(\varphi' - \sqrt{2V(\varphi)}\Bigr)^2 + \int_{-v}^{+v}\sqrt{2V(\varphi)}\mathrm{d}\varphi \\
&= \int_{-\infty}^{+\infty}\mathrm{d}x \tfrac{1}{2}\Bigl(\varphi' - \sqrt{2V(\varphi)}\Bigr)^2 + \tfrac{2}{3}(m^2/\lambda)m.\tag{92.5}
\end{align}
  Since the first term in eq. (92.5) is positive, the minimum possible energy is $M \equiv \tfrac{2}{3}(m^2/\lambda)m$; this is much larger than the particle mass $m$ if the theory is weakly coupled ($\lambda \ll m^2$). Requiring the first term in eq. (92.5) to vanish yields $\varphi' = \sqrt{2V(\varphi)}$, which is easily integrated to get
  $$\varphi(x) = v\tanh\Bigl(\tfrac{1}{2}m(x - x_0)\Bigr),\tag{92.6}$$

How do they get the solution (92.6) where the energy E is missing?
I want to get the equation (92.6) from equation(92.5), can you do some process? 
 A: There are two aspects to this question: a) where do we get the requirement $\varphi' = \sqrt{2 V(\varphi)}$, and b) how do we get 92.6 from that.
a) We're looking for a solution $\varphi(x)$ that minimizes $E$ - we will also want it to be the solution that limits to $\varphi \to \pm v$ as $x\to \pm \infty$.  Eq. 92.5 shows that the energy is a constant term independent of $\varphi$ plus the integral of a positive term (as DJBunk noted above).  The term
$$\int dx \left(\varphi'(x)-\sqrt{2 V(\varphi)}\right)^2 \ge 0$$
since the field is real.  Then its minimum is zero, which occurs if $\varphi'(x)-\sqrt{2 V(\varphi)} = 0$ everywhere.  
Srednicki defines $V(\varphi) = \frac{\lambda}{8} (\varphi^2 - v^2)^2$ and $m = \sqrt{\lambda} v$.  
b) We can tell that 92.6 is right by plugging it into $\varphi'(x)-\sqrt{2 V(\varphi)} = 0$.  This just takes a bit of algebra and the hyperbolic trig identities - the only hard part is to note that since $|\varphi(x)| \le v$, $\sqrt{ (\varphi^2-v^2)^2} = |\varphi^2-v^2| = v^2-\varphi^2$.
Here's how you get 92.6 if you don't already know the answer:  Plugging in $V(\varphi) = \frac{\lambda}{8} (\varphi^2 - v^2)^2$ to $\varphi'(x)=\sqrt{2 V(\varphi)}$, we get 
$$ \varphi'(x) = \frac{\sqrt{\lambda}}{2} \sqrt{ (\varphi^2-v^2)^2} = |\varphi^2-v^2|$$
If we define $\varphi = v \chi$, we get 
$$\chi' = \frac{m}{2} |(\chi^2-1)|$$
The solution we are looking for limits to $\chi \to \pm 1$ as $x\to\pm \infty$; I will assume that these are the maxima and minima of the function, i.e. $|(\chi^2-1)| = 1-\chi^2$.  I'm not sure if this assumption can be avoided cleverly - right now, we can just think of it as a guess, and see if there is a solution that satisfies it.  $\chi' = \frac{m}{2} (\chi^2-1)$ is separable and can be integrated directly:
$$\int \frac{d\chi}{1-\chi^2} = \frac{m}{2} \int dx$$
so
$$\tanh^{-1} \chi = \frac{m}{2} x + \textrm{const.}$$
choosing the constant to be $-\frac{m}{2}x_0$ gives $\chi = \tanh\left[\frac{m}{2} (x-x_0)\right]$ or $\varphi(x) = v\tanh\left[\frac{m}{2} (x-x_0)\right]$, which is 92.6
