If $∫(\sqrt{x+1}+(\sqrt{x-1})^2dx=ax^2+βx(\sqrt{x^2-1}+γlog|x+\sqrt{x^2-1}|+C$, then value of $α+β+2γ$ is:

0

The correct answer is Option (2) → 0

$∫(\sqrt{x+1}+(\sqrt{x-1})^2dx=ax^2+βx(\sqrt{x^2-1}+γ\log|x+\sqrt{x^2-1}|+C$

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

$=2\left[\frac{x^2}{2}+\frac{x}{2}\sqrt{x^2-1}-\frac{1}{2}\log|x+\sqrt{x^2-1}|\right]+C$

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

Matching with the given equation,

$α=1,β=1,γ=-1$

$∴α+β+2γ=1+1-2=0$