Sustitución: $u = \sqrt{x + 1} \implies x = u^2 - 1, dx = 2u \, du$.
Factorizamos $x^2 - 4 = (x - 2)(x + 2)$.
En términos de $u$:
$x - 2 = u^2 - 1 - 2 = u^2 - 3$
$x + 2 = u^2 - 1 + 2 = u^2 + 1$
Sustituyendo:
$$ \int \frac{2u \, du}{(u^2 - 3)(u^2 + 1)u} = 2 \int \frac{du}{(u^2 - 3)(u^2 + 1)} $$
Fracciones parciales para $\frac{1}{(u^2 - 3)(u^2 + 1)}$:
$$ \frac{1}{(u^2 - 3)(u^2 + 1)} = \frac{A}{u^2 - 3} + \frac{B}{u^2 + 1} \implies 1 = A(u^2 + 1) + B(u^2 - 3) $$
Si $u^2 = 3 \implies 1 = 4A \implies A = 1/4$.
Si $u^2 = -1 \implies 1 = -4B \implies B = -1/4$.
Integral:
$$ 2 \left[ \frac{1}{4} \int \frac{du}{u^2 - 3} - \frac{1}{4} \int \frac{du}{u^2 + 1} \right] = \frac{1}{2} \left[ \frac{1}{2\sqrt{3}} \ln \left| \frac{u - \sqrt{3}}{u + \sqrt{3}} \right| - \arctan u \right] + C $$
Sustituyendo $u = \sqrt{x + 1}$:
$$ \frac{1}{4\sqrt{3}} \ln \left| \frac{\sqrt{x + 1} - \sqrt{3}}{\sqrt{x + 1} + \sqrt{3}} \right| - \frac{1}{2} \arctan(\sqrt{x + 1}) + C $$
$$ \boxed{\frac{\sqrt{3}}{12} \ln \left| \frac{\sqrt{x + 1} - \sqrt{3}}{\sqrt{x + 1} + \sqrt{3}} \right| - \frac{1}{2} \arctan(\sqrt{x + 1}) + C} $$