Answer
$$2\sqrt x {\cos ^{ - 1}}\sqrt x - 2\sqrt {1 - x} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\cos }^{ - 1}}\sqrt x }}{{\sqrt x }}} dx \cr
& {\text{substitute }}u = \sqrt x ,{\text{ }}du = \frac{1}{{2\sqrt x }}dx,{\text{ 2}}du = \frac{1}{{\sqrt x }}dx \cr
& \int {\frac{{{{\cos }^{ - 1}}\sqrt x }}{{\sqrt x }}} dx = \int {{{\cos }^{ - 1}}u} \left( {2du} \right) \cr
& = 2\int {{{\cos }^{ - 1}}u} du \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}17 \cr
& \int {{{\cos }^{ - 1}}u} du = u{\cos ^{ - 1}}u - \sqrt {1 - {u^2}} + C \cr
& = 2\int {{{\cos }^{ - 1}}u} du = 2u{\cos ^{ - 1}}u - 2\sqrt {1 - {u^2}} + C \cr
& u = \sqrt x \cr
& = 2\sqrt x {\cos ^{ - 1}}\sqrt x - 2\sqrt {1 - {{\left( {\sqrt x } \right)}^2}} + C \cr
& = 2\sqrt x {\cos ^{ - 1}}\sqrt x - 2\sqrt {1 - x} + C \cr} $$
