Question

If $0<\alpha <1$ and $I={\int }_{-1}^1\frac{dx}{\sqrt{1-2\alpha x+{\alpha }^2}}$ then $I$ equals

• A. $1/\alpha$
• B. $2/\alpha$
• C. $3/\alpha$
• D. none of these

Answer 1

The correct option is D none of these

$I={\int }_{-1}^1(1-2\alpha x+{\alpha }^2{)}^{-1/2}dx$
$={\frac{2(1-2\alpha x+{\alpha }^2{)}^{1/2}}{-2\alpha }]}_{-1}^1$
$=-\frac{1}{\alpha }\left[\right(\left[\right(1-2\alpha +{\alpha }^2{)}^{1/2}-(1+2\alpha +{\alpha }^2{)}^{1/2}]$

$=-\frac{1}{\alpha }[\sqrt{(\alpha -1{)}^2}-\sqrt{(\alpha +1{)}^2}]$

$=-\frac{1}{\alpha }\left[\right(1-\alpha )-(1+\alpha \left)\right]\dots (\because 0<\alpha <1)$

$=2$
Ans: D