If ∫\((x +\sqrt { x^2 -1})^2\)dx =α . x+βx³ +$\gamma$(x² - 1)\(^{\frac{3}{2}}\)+ C, where C is arbitrary constant, then the value of 3(α + β + γ) is

1

\(∫(x +\sqrt { x^2 -1})^2\)dx =α . x+βx³ +$\gamma$(x² - 1)\(^{\frac{3}{2}}\)+ C

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

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

$=\frac{2}{3}x^3-x+\frac{2}{3}.(x^2-1)^{3/2}+C$

Since $\int 2x.\sqrt{x^2-1}dx$

Let $x^2-1=t⇒dt=2xdx$

$\int 2x.\sqrt{x^2-1}dx=\int \sqrt{t}.dt=\frac{t^{3/2}}{3/2}+C$

$=\frac{2}{3}.t^{3/2}+C=\frac{2}{3}(x^2-1)^{3/2}+C$

Comparing  $α =-1, β=2/3, γ=2/3$

$⇒ 3(α + β+ γ) = 3(-1+\frac{2}{3}+\frac{2}{3})=3×\frac{1}{3}=1$

Correct option is (4).