Answer
$$2\ln \left( {\sqrt x + \sqrt {1 + x} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt x \sqrt {1 + x} }}} dx{\text{ }} \cr
& {\text{Let }}{u^2} = x,{\text{ 2}}udu = dx \cr
& {\text{Substitute}} \cr
& \int {\frac{1}{{\sqrt x \sqrt {1 + x} }}} dx{\text{ }} = \int {\frac{1}{{\sqrt {{u^2}} \sqrt {1 + {u^2}} }}\left( {2udu} \right)} {\text{ }} \cr
& = \int {\frac{{2u}}{{u\sqrt {1 + {u^2}} }}du} {\text{ }} \cr
& = 2\int {\frac{1}{{\sqrt {1 + {u^2}} }}du} {\text{ }} \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }} = } \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& = 2\ln \left( {u + \sqrt {1 + {u^2}} } \right) + C \cr
& {\text{Write in terms of }}x \cr
& = 2\ln \left( {\sqrt x + \sqrt {1 + x} } \right) + C \cr} $$
