1. Análisis de sustitución
Dada la forma $a^2 + u^2$, usamos sustitución trigonométrica.
Sea $\sqrt{3}x = \sqrt{2} \tan \theta$, entonces:
$x = \sqrt{\frac{2}{3}} \tan \theta \implies dx = \sqrt{\frac{2}{3}} \sec^2 \theta d\theta$.
Además, $2 + 3x^2 = 2(1 + \tan^2 \theta) = 2 \sec^2 \theta$.

2. Sustitución en la integral
$$ \begin{aligned} I &= \int \frac{\sqrt{\frac{2}{3}} \sec^2 \theta d\theta}{\left(\sqrt{\frac{2}{3}} \tan \theta\right)^2 (2 \sec^2 \theta)^{5/2}} \\ I &= \int \frac{\sqrt{\frac{2}{3}} \sec^2 \theta}{\frac{2}{3} \tan^2 \theta \cdot 2^{5/2} \sec^5 \theta} d\theta = \frac{\sqrt{2/3} \cdot 3}{2 \cdot 4\sqrt{2}} \int \frac{1}{\tan^2 \theta \sec^3 \theta} d\theta \\ I &= \frac{\sqrt{2}/\sqrt{3} \cdot 3}{8\sqrt{2}} \int \frac{\cos^2 \theta}{\sin^2 \theta} \cos^3 \theta d\theta = \frac{\sqrt{3}}{8} \int \frac{\cos^5 \theta}{\sin^2 \theta} d\theta \end{aligned} $$

3. Resolución de la integral de senos y cosenos
$\int \frac{(1 - \sin^2 \theta)^2 \cos \theta}{\sin^2 \theta} d\theta$. Sea $u = \sin \theta, du = \cos \theta d\theta$:
$$ \int \frac{(1 - u^2)^2}{u^2} du = \int \frac{1 - 2u^2 + u^4}{u^2} du = \int (u^{-2} - 2 + u^2) du = -\frac{1}{u} - 2u + \frac{u^3}{3} $$

4. Retorno a la variable $x$
Usando el triángulo para $\tan \theta = \frac{\sqrt{3}x}{\sqrt{2}}$, tenemos $\sin \theta = \frac{\sqrt{3}x}{\sqrt{2+3x^2}}$.
Sustituyendo estos valores se llega a la expresión final simplificada.
$$ \boxed{I = \frac{\sqrt{3}}{8} \left[ -\frac{\sqrt{2+3x^2}}{\sqrt{3}x} - 2\frac{\sqrt{3}x}{\sqrt{2+3x^2}} + \frac{1}{3}\left(\frac{\sqrt{3}x}{\sqrt{2+3x^2}}\right)^3 \right] + C} $$