= Solution
Measure intermediate time $u$ from $t'$. The <classical path> with the fixed endpoints is
$$
x_c(u)=x'+\frac{X}{T}u+\frac{\alpha}{2m}u(T-u),\qquad 0\leq u\leq T.
$$
It satisfies $m\ddot x_c=-\alpha$. Direct integration of $m\dot x_c^2/2-\alpha x_c$ gives
$$
\boxed{S_c=\frac{mX^2}{2T}-\frac{\alpha T(x+x')}{2}-\frac{\alpha^2T^3}{24m}}.
$$
For an arbitrary path write $x=x_c+\eta$ with $\eta(0)=\eta(T)=0$. Integrating the cross term by parts gives $-\int(m\ddot x_c+\alpha)\eta\,du=0$. Thus $S[x]=S_c+\int m\dot\eta^2du/2$ exactly. The fluctuation <path integral> is the free endpoint-fixed one, so its <determinant> supplies $\sqrt{m/(2\pi i\hbar T)}$. Multiplication by $\theta(T)e^{iS_c/\hbar}$ reproduces the previous kernel. This exact <classical-path factorization of a quadratic path integral> is possible because a linear potential introduces no additional quadratic fluctuation term.
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