Let $\int \frac{dx}{\left(\sqrt{x} - \sqrt{x-1}\right)^2} = \alpha u(x) + \beta v(x) + C$, where $u(x) = x^2 - x + \left(x - \frac{1}{2}\right)\sqrt{x^2 - x}$ and $v(x) = \log\left|x - \frac{1}{2} + \sqrt{x^2 - x}\right|$. The value of $\alpha + \beta$ is :

Rationalize: $\frac{1}{(\sqrt{x} - \sqrt{x-1})^2} = (\sqrt{x} + \sqrt{x-1})^2 = 2x - 1 + 2\sqrt{x^2 - x}$. Integrating: $x^2 - x + 2\int\sqrt{x^2-x}\,dx$. With $x^2-x = (x-1/2)^2 - 1/4$, $2\int\sqrt{x^2-x}\,dx = (x-1/2)\sqrt{x^2-x} - \frac{1}{4}\log|x-1/2 + \sqrt{x^2-x}|$. So total $= u(x) - \frac{1}{4}v(x) + C$. Hence $\alpha = 1$, $\beta = -\frac{1}{4}$, $\alpha + \beta = \frac{3}{4}$.