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

$α=β$

The correct answer is Option (3) → $α=β$

Given integral: $I = \int \frac{x}{x - \sqrt{x^2 - 4}} \, dx$

Rationalize the denominator by multiplying numerator and denominator by $x + \sqrt{x^2 - 4}$:

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

So integral becomes:

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

Compute each part:

$\frac{1}{4} \int x^2 dx = \frac{x^3}{12}$

Let $u = x^2 - 4 \Rightarrow du = 2x dx \Rightarrow x dx = \frac{du}{2}$

$\frac{1}{4} \int x \sqrt{x^2 - 4} dx = \frac{1}{4} \int \frac{\sqrt{u}}{2} du = \frac{1}{8} \int u^{1/2} du = \frac{1}{8} \cdot \frac{2}{3} u^{3/2} = \frac{(x^2 - 4)^{3/2}}{12}$

Therefore:

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

Answer: $\alpha = \frac{1}{12}, \beta = \frac{1}{12}, \gamma = \text{constant}$