Answer
$$2\ln \left( {\sqrt {1 + x} + \sqrt x } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt x \cdot \sqrt {1 + x} }}} dx \cr
& {\text{Write the integrand as}} \cr
& \int {\frac{1}{{\sqrt x \cdot \sqrt {1 + x} }}} dx = \int {\frac{1}{{\sqrt x \cdot \sqrt {1 + {{\left( {\sqrt x } \right)}^2}} }}} dx \cr
& {\text{Use the substitution method}}{\text{, }} \cr
& \,\,\,\,u = \sqrt x ,\,\,\,\,du = \frac{1}{{2\sqrt x }}dx \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\frac{1}{{\sqrt x \cdot \sqrt {1 + {{\left( {\sqrt x } \right)}^2}} }}} dx = \int {\frac{{2du}}{{\sqrt {1 + {u^2}} }}} \cr
& = 2\int {\frac{{du}}{{\sqrt {1 + {u^2}} }}} \cr
& {\text{Use the trigonometric substitution }}u = \tan \theta ,\,\,\,du = {\sec ^2}\theta d\theta \cr
& = 2\int {\frac{{du}}{{\sqrt {1 + {u^2}} }}} = 2\int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}} \cr
& = 2\int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \cr
& = 2\int {\sec \theta } d\theta \cr
& \cr
& {\text{Integrate}} \cr
& = 2\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{Where sec}}\theta = \sqrt {1 + {u^2}} {\text{ and tan}}\theta = u \cr
& = 2\ln \left| {\sqrt {1 + {u^2}} + u} \right| + C \cr
& {\text{Write the integrand in terms of }}x,{\text{ substitute }}\sqrt x {\text{ for }}u \cr
& = 2\ln \left| {\sqrt {1 + {{\left( {\sqrt x } \right)}^2}} + \sqrt x } \right| + C \cr
& = 2\ln \left( {\sqrt {1 + x} + \sqrt x } \right) + C \cr} $$
