1. Completar el cuadrado:
$x^2 + 2x - 4 = (x + 1)^2 - 5$.
Sustituyendo $u = x + 1$, $du = dx$:
$$ \int \frac{du}{u^3 \sqrt{u^2 - 5}} $$

2. Sustitución trigonométrica:
Sea $u = \sqrt{5}\sec(\theta)$, $du = \sqrt{5}\sec(\theta)\tan(\theta) d\theta$:
$$ \int \frac{\sqrt{5}\sec(\theta)\tan(\theta) d\theta}{5\sqrt{5}\sec^3(\theta) \sqrt{5}\tan(\theta)} = \frac{1}{5\sqrt{5}} \int \frac{1}{\sec^2(\theta)} d\theta $$
$$ = \frac{1}{5\sqrt{5}} \int \cos^2(\theta) d\theta = \frac{1}{5\sqrt{5}} \int \frac{1 + \cos(2\theta)}{2} d\theta $$
$$ = \frac{1}{10\sqrt{5}} \left( \theta + \frac{\sin(2\theta)}{2} \right) + C = \frac{1}{10\sqrt{5}} (\theta + \sin(\theta)\cos(\theta)) + C $$
Como $\sec(\theta) = \frac{u}{\sqrt{5}}$, entonces $\cos(\theta) = \frac{\sqrt{5}}{u}$ y $\sin(\theta) = \frac{\sqrt{u^2-5}}{u}$.
$$ = \frac{1}{10\sqrt{5}} \left( \sec^{-1}\left(\frac{x+1}{\sqrt{5}}\right) + \frac{\sqrt{5}\sqrt{x^2+2x-4}}{(x+1)^2} \right) + C $$
$$ \boxed{\frac{1}{10\sqrt{5}}\sec^{-1}\left(\frac{x+1}{\sqrt{5}}\right) + \frac{\sqrt{x^2+2x-4}}{10(x+1)^2} + C} $$