Question:

Let $I = \int \frac{dx}{(x-\alpha)\sqrt{(x-\alpha)(x-\beta)}} = \int \frac{dx}{(x-\alpha)\sqrt{(x-\alpha)^2 \frac{x-\beta}{x-\alpha}}} = \int \frac{dx}{(x-\alpha)^2 \sqrt{\frac{x-\beta}{x-\alpha}}}$. Let $t = \sqrt{\frac{x-\beta}{x-\alpha}}$. Then $t^2 = \frac{x-\beta}{x-\alpha} = \frac{x-\alpha+\alpha-\beta}{x-\alpha} = 1 + \frac{\alpha-\beta}{x-\alpha}$. So $\frac{\alpha-\beta}{x-\alpha} = t^2-1$, which means $x-\alpha = \frac{\alpha-\beta}{t^2-1}$. Differentiating, $dx = \frac{-(\alpha-\beta) \cdot 2t}{(t^2-1)^2} dt$. Substituting these into the integral: $I = \int \frac{1}{(\frac{\alpha-\beta}{t^2-1})^2 \cdot t} \cdot \frac{-2t(\alpha-\beta)}{(t^2-1)^2} dt = \int \frac{-2(\alpha-\beta)(t^2-1)^2}{(\alpha-\beta)^2 t} \cdot \frac{t}{(t^2-1)^2} dt = \int \frac{-2}{\alpha-\beta} dt = \frac{-2}{\alpha-\beta} t + C = \frac{-2}{\alpha-\beta} \sqrt{\frac{x-\beta}{x-\alpha}} + C$.