Question

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

• A. ${\cos }^{-1}\frac{1}{|x|}-\frac{\sqrt{x^2-1}}{x}+c$
• B. ${\cos }^{-1}\frac{1}{|x|}+\frac{\sqrt{x^2-1}}{x}+c$
• C. ${\sin }^{-2}\frac{1}{|x|}-\frac{\sqrt{x^2-1}}{x}+c$
• D. ${\sin }^{-1}\frac{1}{|x|}+\frac{\sqrt{x^2-1}}{x}+c$

Answer 1

Answer: (A)

${\cos }^{-1}\frac{1}{|x|}-\frac{\sqrt{x^2-1}}{x}+c$

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

$\frac{1}{x}=t-\frac{1}{x^2}dx=dt$

$\frac{1}{x^2}dx=-dt$

$-\int \sqrt{\frac{1-t}{1+t}}dt$

$t=cos2\theta$

$dt=-25in2\theta -d\theta$

$-\int \sqrt{\frac{2si{n}^2\theta }{2co{s}^2\theta }}-2\left(sin2\theta \right)d\theta$

$=+2\int \tan \theta sin2\theta d\theta$

$4\int si{n}^2\theta d\theta$

$4\int {\left(\frac{1-cos2\theta }{2}\right)}^{d\theta }+C$

$4[\frac{1}{2}\theta -\frac{1}{4}sin2\theta ]+c$

$2\theta -sin2\theta +c$

$\int \frac{1}{x^2}\sqrt{\frac{x-1}{x+1}}dx=co{s}^{-1}\left|\frac{1}{x}\right|-\frac{\sqrt{x^2-1}}{x}+c$