Question

If $I=\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}dx$ then $I$ equals to :

• A. $\log \left|x\right|+\log |1+\sqrt{1-x^2}|+{\sin }^{-1}\sqrt{x}+C$
• B. $\log \left|x\right|-\log |1-\sqrt{1-x^2}|+{\tan }^{-1}\sqrt{x}+C$
• C. $\log \left|x\right|-\log |1+\sqrt{1-x^2}|-{\sin }^{-1}x+C$
• D. $\log |1+\sqrt{1-x^2}|-\log \left|x\right|+{\cos }^{-1}x+C$

Answer 1

Answer: (C)

$\log \left|x\right|-\log |1+\sqrt{1-x^2}|-{\sin }^{-1}x+C$

Let $I=\int \frac{1}{x}\sqrt{\frac{1-x}{1+x}}dx=\int \frac{1}{x}\times \frac{1-x}{\sqrt{1-x^2}}dx$
$=\int \frac{dx}{x\sqrt{1-x^2}}-\int \frac{dx}{\sqrt{1-x^2}}$
Substitute $t=\frac{1}{x}$ in $1^{st}$ integrals
$I=\int \frac{-\frac{1}{t^2}}{\frac{1}{t}\sqrt{1-\frac{1}{t^2}}}dt-{\sin }^{-1}x$
$=-\int \frac{dt}{\sqrt{t^2-1}}-{\sin }^{-1}x=-\log (t+\sqrt{t^2-1})-{\sin }^{-1}x+c$
$=-\log (\frac{1}{x}+\frac{\sqrt{1-x^2}}{x})-{\sin }^{-1}x+c$
$=\log |x|-\log |(1+\sqrt{1-x^2})|-{\sin }^{-1}x+c$