1. Estrategia
Utilizamos la sustitución $t = \frac{x-3}{x-2}$.
Entonces $x-2 = \frac{1}{1-t}$ y $dx = \frac{-1}{(1-t)^2} dt$. Además $x-3 = t(x-2) = \frac{t}{1-t}$.

2. Sustitución en la integral
$$ \begin{aligned} I &= \int \frac{\frac{-1}{(1-t)^2} dt}{\left(\frac{t}{1-t}\right)^4 \left(\frac{1}{1-t}\right)^5} = \int \frac{-dt}{(1-t)^2 \cdot \frac{t^4}{(1-t)^4} \cdot \frac{1}{(1-t)^5}} \\ &= -\int \frac{(1-t)^7}{t^4} dt \end{aligned} $$

3. Desarrollo e integración
Expandiendo $(1-t)^7$:
$$ (1-t)^7 = 1 - 7t + 21t^2 - 35t^3 + 35t^4 - 21t^5 + 7t^6 - t^7 $$
Dividiendo entre $t^4$:
$$ I = -\int (t^{-4} - 7t^{-3} + 21t^{-2} - 35t^{-1} + 35 - 21t + 7t^2 - t^3) dt $$
$$ I = -\left[ \frac{t^{-3}}{-3} - \frac{7t^{-2}}{-2} + \frac{21t^{-1}}{-1} - 35\ln|t| + 35t - \frac{21t^2}{2} + \frac{7t^3}{3} - \frac{t^4}{4} \right] + C $$

4. Resultado final
Sustituyendo $t = \frac{x-3}{x-2}$:
$$ \boxed{I = \frac{1}{3}\left(\frac{x-2}{x-3}\right)^3 - \frac{7}{2}\left(\frac{x-2}{x-3}\right)^2 + 21\left(\frac{x-2}{x-3}\right) + 35\ln\left|\frac{x-3}{x-2}\right| - 35t + \frac{21t^2}{2} - \frac{7t^3}{3} + \frac{t^4}{4} + C} $$
donde $t = \frac{x-3}{x-2}$.