Question

If $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx=A\sqrt{1-x}+B{\sin }^{-1}\sqrt{x}+C\sqrt{x-x^2}+D,$ where $A+B+C=$

• A. $-2$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $-2$

$=-2\sqrt{1-x}-{\sin }^{-1}\sqrt{x}+\sqrt{x(1-x)}+c$

$=-2\sqrt{1-x}-{\sin }^{-1}\sqrt{x}+\sqrt{x-x^2}+c$

by comparing with $=A\sqrt{1-x}+B{\sin }^{-1}\sqrt{x}+C\sqrt{x-x^2}+D$

$A=-2,B=-1,C=1$

$\therefore A+B+C=-2-1+1=-2$