= Solution
For <potential flow> $\mathbf u=\nabla\varphi$ in an inviscid fluid of constant density under the <conservative force> per unit mass $-\nabla\chi$, the unsteady <Bernoulli equation> is
$$
\frac p\rho+\partial_t\varphi+\frac12|\nabla\varphi|^2+\chi=C(t).
$$
It follows by inserting the <velocity potential> into the <Euler equations> and integrating the resulting spatial gradient. A time-only change of the <velocity potential> changes $C(t)$ without altering the <velocity field>.
For spherical motion in an unbounded liquid, incompressibility gives $r^2u_r=A(t)$; the kinematic condition $u_r(R)=\dot R$ gives $A=R^2\dot R$. Choose $\varphi=-R^2\dot R/r$, vanishing at infinity. To obtain the displayed bubble relation, take no spatial body-force contribution, neglect <viscosity> and <surface tension>, and equate the liquid-side interface pressure to $p_b$. With the far-field pressure $p_\infty$, <Bernoulli equation> evaluated at the surface uses the partial derivative at fixed $r$:
$$
\left.\partial_t\varphi\right|_{r=R}
=-2\dot R^2-R\ddot R,\qquad
\left.|\mathbf u|^2\right|_{r=R}=\dot R^2.
$$
Hence the <Rayleigh equation for an inviscid spherical bubble> is
$$
\boxed{p_b-p_\infty=\rho\left(R\ddot R+\frac32\dot R^2\right)}.
$$
The exterior <kinetic energy> is
$$
K=\frac\rho2\int_R^\infty4\pi r^2\frac{R^4\dot R^2}{r^4}\,dr
=\boxed{2\pi\rho R^3\dot R^2}.
$$
Differentiating it and using $\dot V=4\pi R^2\dot R$ gives the <pressure-work energy balance for a spherical bubble>:
$$
\dot K=4\pi\rho R^2\dot R\left(R\ddot R+\frac32\dot R^2\right)
=\boxed{(p_b-p_\infty)\dot V}.
$$
This derivation also holds at a turning point $\dot R=0$, without dividing by $\dot R$.
For constant $p_\infty$ and the stated inverse-volume bubble pressure, integrate this exact time derivative:
$$
K-K_0=p_\infty\int_{V_0}^V\left(\frac{V_0}{s}-1\right)\,ds
=\boxed{p_\infty\left[V_0\log\!\left(\frac V{V_0}\right)-V+V_0\right]}.
$$
The relation is valid on physically accessible portions of the motion, where the resulting <kinetic energy> is nonnegative. In particular $\log q\le q-1$ makes the bracket nonpositive, as expected for the restoring pressure about $V_0$.
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