I stumbled upon a derivation that works, although doesn't involve the steps mentioned in the book:
$\frac{dL(p + t\omega)}{dt} = -(\sigma_{maj} - \sigma_n(p + t\omega))L(p + t\omega)$.
$T_r(p + t\omega \to p') = \frac{L(p')}{L(p + t\omega)}$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = L(p')\frac{d}{dt}(L(p + t\omega)^{-1})$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = -L(p')L(p + t\omega)^{-2}\frac{d}{dt}(L(p + t\omega))$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = -\frac{L(p')}{L(p + t\omega)^{2}}\frac{dL(p + t\omega)}{dt}$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = -\frac{L(p')}{L(p + t\omega)^{2}}(-(\sigma_{maj} - \sigma_n(p + t\omega))L(p + t\omega))$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = \frac{L(p')}{L(p + t\omega)}(\sigma_{maj} - \sigma_n(p + t\omega))$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) = T_r(p + t\omega \to p')(\sigma_{maj} - \sigma_n(p + t\omega))$,
$\frac{d}{dt}(T_r(p + t\omega \to p')) - \sigma_{maj}T_r(p + t\omega \to p') = - \sigma_n(p + t\omega)T_r(p + t\omega \to p')$.
Multiply by $e^{-\sigma_{maj}t}$:
$e^{-\sigma_{maj}t}\frac{d}{dt}(T_r(p + t\omega \to p')) - \sigma_{maj}e^{-\sigma_{maj}t}T_r(p + t\omega \to p') = - e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p')$
Applying the product rule:
$\frac{d}{dt}(e^{-\sigma_{maj}t}T_r(p + t\omega \to p')) = - e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p')$,
$\int_0^d d(e^{-\sigma_{maj}t}T_r(p + t\omega \to p')) = - \int_0^d e^{-\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$,
$e^{-\sigma_{maj}d} - T_r(p \to p') = - \int_0^d e^{ \sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$$e^{-\sigma_{maj}d} - T_r(p \to p') = - \int_0^d e^{ -\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$,
$T_r(p \to p') = e^{-\sigma_{maj}d} + \int_0^d e^{ \sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$$T_r(p \to p') = e^{-\sigma_{maj}d} + \int_0^d e^{ -\sigma_{maj}t} \sigma_n(p + t\omega)T_r(p + t\omega \to p') dt$.