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?