integral

\int \frac{dx}{ 1+ \sqrt{x}.\sqrt{x}.\sqrt{1- x}}

\sqrt{x} = t

on differentiating

\frac{1}{2\sqrt{x}} dx = dt

now integral becomes

2 \int \frac{dt}{( 1 + t ) . \sqrt{1- t^2 }}

now put ----1 + t = \frac{1}{z} \Rightarrow t = \frac{1}{z} -1 = \frac{1- z}{z}

dt = \frac{-1}{z^2 }

integral reduces to

- 2\int \frac{dz }{\sqrt{2z - 1}} = - 2 \sqrt{2z - 1}

now put the value of z we get answer as

\frac{2(\sqrt{x}-1)}{\sqrt{1 - x}} + c

hope this helps but hari sir 's method is very nice and ShOrT

yup ur ans is also correct [1], nice solution omkar!!