If ∫(√(x+1)+√(x-1))^2dx=ax^2+βx(√(x^2-1)+γlog|x+√(x^2-1)|+C, then val…

$\int (\sqrt{x+1}+\sqrt{x-1})^2 dx$

$= \int \left[(x+1)+(x-1)+2\sqrt{x^2-1}\right] dx$

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

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

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

$\int \sqrt{x^2-1} \, dx = \frac{x}{2}\sqrt{x^2-1} - \frac{1}{2}\ln|x+\sqrt{x^2-1}|$

$\Rightarrow 2\int \sqrt{x^2-1} dx = x\sqrt{x^2-1} - \ln|x+\sqrt{x^2-1}|$

$\alpha = 1,\ \beta = 1,\ \gamma = -1$

$\alpha + \beta + 2\gamma = 1 + 1 + 2(-1) = 0$

The value is $0$.