Question

The value of the integer $I={\int }_1^{\infty }\frac{(x^2-x)}{x^3\sqrt{(x^2-1)}}dx$ is

• A. $0$
• B. $2/3$
• C. $4/3$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

Let, $x=\sec t⟹dx=\sec t\tan tdt$

As $x:1\to \infty ,\sec t:0\to \pi /2$

$I={\int }_1^{\infty }\frac{(x^2-x)}{x^3\sqrt{x^2-1}}dx$

$I={\int }_1^{\infty }\frac{(x-1)}{x^2\sqrt{x^2-1}}dx$

$I={\int }_0^{\pi /2}\frac{(\sec t-1)}{{\sec }^{2}t\sqrt{{\sec }^{2}t-1}}\sec t\tan tdt$

$I={\int }_0^{\pi /2}\frac{(\sec t-1)}{{\sec }^{2}t.\tan t}\sec t\tan tdt$

$I={\int }_0^{\pi /2}\frac{(\sec t-1)}{\sec t}dt$

$I={\int }_0^{\pi /2}(1-\cos t)dt$

$I=[t-\sin t{]}_0^{\pi /2}$

$I=\frac{\pi }{2}-1$(Ans)

Option D is correct.