= Solution
Conditioned on $f$, the additive-noise trajectory has the <Onsager–Machlup functional>
$$
P[x\mid f]\propto
\exp\left[
-\frac1{2C^2}\int_0^T
(\dot x+V'(x)-f)^2dt
\right].
$$
Under time reversal, $\dot x$ changes sign while $x$ and the active force $f$ are even. Subtracting the forward and backward conditional actions gives
$$
\log\frac{P_F[x\mid f]}{P_B[x\mid f]}
=\frac2{C^2}\left[
V(x_0)-V(x_T)
+\int_0^T\dot x\,f\,dt
\right].
$$
The corresponding ratio for an Ornstein–Uhlenbeck path conditioned on its initial endpoint is
$$
\log\frac{P_F[f]}{P_B[f]}
=\frac{\alpha}{c^2}
\left[f(0)^2-f(T)^2\right].
$$
Consequently
$$
\boxed{
\log\frac{P_F[f,x]}{P_B[f,x]}
=\Delta U[f,x]
+\frac2{C^2}\int_0^T\dot x(t)f(t)\,dt
},
$$
where
$$
\boxed{
\Delta U[f,x]
=\frac{\alpha}{c^2}[f(0)^2-f(T)^2]
+\frac2{C^2}[V(x_0)-V(x_T)]
}.
$$
If stationary endpoint densities are included in the path measures, their ratio cancels the Ornstein–Uhlenbeck boundary term; the displayed convention is the endpoint-conditioned path probability used in the calculation.
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