Answer
$$\arcsin \sqrt x + \sqrt {x\left( {1 - x} \right)} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 - x} }}{{\sqrt x }}} dx \cr
& {\text{Rewrite}} \cr
& \int {\frac{{\sqrt {1 - x} }}{{\sqrt x }}} dx = \int {\frac{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}{{\sqrt x }}} dx \cr
& {\text{Let }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ }}dx = 2\sqrt x du \cr
& \int {\frac{{\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} }}{{\sqrt x }}} dx = \int {\frac{{\sqrt {1 - {u^2}} }}{{\sqrt x }}} \left( {2\sqrt x } \right)du \cr
& = 2\int {\sqrt {1 - {u^2}} } du \cr
& {\text{Use the special integration formula }}\left( {{\text{Theorem 8}}{\text{.2}}} \right) \cr
& \int {\sqrt {{a^2} - {u^2}} du = \frac{1}{2}\left( {{a^2}\arcsin \frac{u}{a} + u\sqrt {{a^2} - {u^2}} } \right) + C} \cr
& 2\int {\sqrt {1 - {u^2}} } du = {\left( 1 \right)^2}\arcsin \frac{{\sqrt x }}{1} + \sqrt x \sqrt {1 - {{\left( {\sqrt x } \right)}^2}} + C \cr
& {\text{Simplifying}} \cr
& \arcsin \sqrt x + \sqrt x \sqrt {1 - x} + C \cr
& \arcsin \sqrt x + \sqrt {x\left( {1 - x} \right)} + C \cr} $$