If $I = \int \frac{x}{x - \sqrt{x^2 - 4}} dx = \alpha x^3 + \beta(x^2 - 4)^{\frac{3}{2}} + \gamma$, where $\gamma$ is constant of integration, then

Multiply both numerator and denominator by $(x + \sqrt{x^2 - 4})$:

$I = \int \frac{x}{x - \sqrt{x^2 - 4}} \cdot \frac{x + \sqrt{x^2 - 4}}{x + \sqrt{x^2 - 4}} dx$

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The denominator simplifies to:

$(x - \sqrt{x^2 - 4})(x + \sqrt{x^2 - 4})$

$= x^2 - (x^2 - 4)$

$= 4$

The numerator becomes $x(x + \sqrt{x^2 - 4})$

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The integral simplifies to:

$I = \int \frac{x(x + \sqrt{x^2 - 4})}{4} dx$

$= \frac{1}{4} \int [x^2 + x\sqrt{x^2 - 4}] dx$

$= \frac{1}{4} \int x^2 dx + \frac{1}{4} \int x\sqrt{x^2 - 4} dx$

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The first integral:

$\int x^2 dx = \frac{x^3}{3}$

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For the second integral, substitute $u = x^2 - 4$, so $x \, dx = \frac{du}{2}$:

$\int x\sqrt{x^2 - 4} dx = \int \sqrt{u} \cdot \frac{du}{2}$

$= \frac{1}{2} \int u^{1/2} du$

$= \frac{1}{2} \cdot \frac{u^{3/2}}{3/2}$

$= \frac{1}{3}u^{3/2}$

$= \frac{1}{3}(x^2 - 4)^{3/2}$

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Combining both integrals:

$I = \frac{1}{4} \cdot \frac{x^3}{3} + \frac{1}{4} \cdot \frac{1}{3}(x^2 - 4)^{3/2} + \gamma$

$= \frac{1}{12}x^3 + \frac{1}{12}(x^2 - 4)^{3/2} + \gamma$

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Comparing with the given form $I = \alpha x^3 + \beta(x^2 - 4)^{\frac{3}{2}} + \gamma$:

$\alpha = \frac{1}{12}$

$\beta = \frac{1}{12}$

Therefore, $\alpha = \beta$