\( \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} d x= \)

(A)
\( \int_{0}^{1} \sqrt{\frac{1+x}{1-x} d x} \)
\( =\int_{0}^{1} \sqrt{\frac{(1+x)^{2}}{(1+x)(1-x)}} d x \)
\( =\int_{0}^{1} \frac{1+x}{\sqrt{1-x^{2}}} d x \)
\( =\int_{0}^{1} \frac{1}{\sqrt{1-x^{2}}+\int_{0}^{1} x d x \sqrt{1-x^{2}}} \)
\( =\left[\sin ^{-1}\right]_{0}{ }^{1}+\int_{0}^{1} \frac{x d x}{\sqrt{1-x^{2}}} \)
\( =\left(\frac{\Pi}{2}-0\right)+\int_{0}^{1} \frac{x d x}{\sqrt{1-x^{2}}} \)
put \( 1-x^{2}=t \Rightarrow-2 x d x=d t \)
\( \Rightarrow x d x=-\frac{d t}{2} \)
at \( x=0 ; t=1 \)
\( x=1 ; t=0 \)
\( \therefore I=\frac{\Pi}{2}+\int_{0}^{1} \frac{-d t}{2 \sqrt{t}} \)
\( =\frac{\Pi}{2}+(-\sqrt{t})_{1}^{0}=\frac{I}{2}+(0+1)=\frac{I}{2}+1 \)