A integration is given, $$M = \int_{- \infty}^{\infty} \left[\frac{1}{2} \left(\frac{d\phi}{dx}\right)^2 + \frac{\lambda}{4}(\phi^2-v^2)^2\right] dx,$$ where $$m=v\sqrt\lambda$$ and $$ \phi(x)= v\tanh\left[ \frac{m}{\sqrt 2} (x-x_0)\right]$$
How this integration gives =$\frac{2 \sqrt2}{3} \frac{m^3}{\lambda}?$
Notice
$$\frac{d\phi}{dx} = \frac{vm}{\sqrt{2}} \frac{1}{\cosh(\frac{m}{\sqrt{2}}(x-x_0))^2} = \frac{vm}{\sqrt{2}}\left(1 - \left(\frac{\phi}{v}\right)^2\right) = \sqrt{\frac{\lambda}{2}}\left(v^2 - \phi^2\right) $$
$M$ can be rewritten as:
$$M = \int_{-\infty}^{\infty}\left(\frac12\sqrt{\frac{\lambda}{2}}+\frac{\lambda}{4}\sqrt{\frac{2}{\lambda}}\right) (v^2 - \phi^2) \frac{d\phi}{dx} dx = \sqrt{\frac{\lambda}{2}}\int_{-v}^{v} (v^2 - \phi^2) d\phi\\ = \sqrt{\frac{\lambda}{2}}\left(\frac43 v^3\right) = \frac{2\sqrt{2\lambda}}{3}\left(\frac{m}{\sqrt{\lambda}}\right)^3 = \frac{2\sqrt{2}}{3}\frac{m^3}{\lambda} $$