$∫\frac{dx}{x-\sqrt{x^2-1}}= αx^2+βx\sqrt{x^2-1}+γlog|x+\sqrt{x^2-1}|+C$, then the value of α + β + γ is:

$\frac{1}{2}$

$∫\frac{dx}{x-\sqrt{x^2-1}}×\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=∫\frac{x+x\sqrt{x^2-1}}{x^2-(x^2-1)}dx$

$∫x+\sqrt{x^2-1}dx=\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2-1}-\frac{1}{2}log|x+\sqrt{x^2-1}|+C$

$=αx^2+βx\sqrt{x^2-1}+γlog|x+\sqrt{x^2-1}|+C$

Comparing, $α=\frac{1}{2},\, β=\frac{1}{2},\, γ=\frac{-1}{2}$

$∴α+β+γ=\frac{1}{2}+\frac{1}{2}-\frac{1}{2}=\frac{1}{2}$