Question

$LetI=\underset{0}{\overset{1}{\int }}\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}dx$ and $J=\underset{0}{\overset{1}{\int }}\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx,$, then correct statement is

• A. $I+J=4$
• B. $I+J=\pi$
• C. $I-J=\pi$
• D. $J=\frac{4-\pi }{2}$

Answer 1

The correct option is A $I+J=4$

$I={\int }_0^1\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}dx$

$J={\int }_0^1\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$

$I+J={\int }_0^1\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx+{\int }_0^1\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}}dx$

$I+J={\int }_0^1(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+\sqrt{\frac{1+\sqrt{x}}{1-\sqrt{x}}})dx$

$I+J={\int }_0^1\left(\frac{1-\sqrt{x}+1+\sqrt{x}}{\sqrt{(1+\sqrt{x})(1-\sqrt{x})}}\right)dx$

$I+J={\int }_0^1\frac{2}{\sqrt{1-x}}dx$

$I+J=2\ast [-2\sqrt{1-x}{]}_0^1$

$I+J=4$