= Solution
For the true <martingale> of part (a), $M_0=u(T,x)$ and $M_T=f(B_T)e^{A_T}$. Taking <expectations> gives the <Feynman-Kac formula with a bounded potential>:
$$
\boxed{u(t,x)=\mathbb E_x\left[f(B_t)\exp\!\left(\int_0^t\frac{ds}{1+B_s^2}\right)\right].}
$$
The right side is finite since its absolute value is at most $e^t\|f\|_\infty$. This also proves uniqueness among <classical solutions> bounded on every finite time slab: applying the formula to their difference with zero initial data gives zero. The positive sign of the potential in the <partial differential equation> explains the positive exponent; changing it to a negative exponent would solve a different equation.
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