Answer
$$\frac{4}{3}\left( {1 + \sqrt {1 + x} } \right)\sqrt {1 + \sqrt {1 + x} {\text{ }}} - 4\sqrt {1 + \sqrt {1 + x} {\text{ }}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {1 + \sqrt {1 + x} } }}} \cr
& {\text{Let }}u = \sqrt {1 + x} {\text{ }} \Rightarrow {\text{ }}du = \frac{1}{{2\sqrt {1 + x} }}dx{\text{ }} \cr
& {\text{ }} \Rightarrow {\text{ }}dx = 2\sqrt {1 + x} du \cr
& {\text{ }} \Rightarrow {\text{ }}dx = 2udu \cr
& {\text{Use the substitution}} \cr
& \int {\frac{{dx}}{{\sqrt {1 + \sqrt {1 + x} } }}} = \int {\frac{{2u}}{{\sqrt {1 + u} }}} du \cr
& {\text{Let }}z = 1 + u{\text{ }} \Rightarrow {\text{ }}dz = du \cr
& \int {\frac{{2u}}{{\sqrt {1 + u} }}} du = \int {\frac{{2\left( {z - 1} \right)}}{{\sqrt z }}} dz \cr
& {\text{ }} = \int {\frac{{2z - 2}}{{\sqrt z }}} dz \cr
& {\text{ }} = \int {\left( {2{z^{1/2}} - 2{z^{ - 1/2}}} \right)} dz \cr
& {\text{Integrating}} \cr
& {\text{ }} = 2\left( {\frac{{{z^{3/2}}}}{{3/2}}} \right) - 2\left( {\frac{{{z^{1/2}}}}{{1/2}}} \right) + C \cr
& {\text{ }} = \frac{4}{3}{z^{3/2}} - 4{z^{1/2}} + C \cr
& {\text{ }} = \frac{4}{3}z\sqrt z - 4\sqrt z + C \cr
& {\text{Back - substitute }}z = 1 + u{\text{ }} \cr
& {\text{ }} = \frac{4}{3}\left( {1 + u} \right)\sqrt {1 + u{\text{ }}} - 4\sqrt {1 + u{\text{ }}} + C \cr
& {\text{Back - substitute }}u = \sqrt {1 + x} \cr
& {\text{ }} = \frac{4}{3}\left( {1 + \sqrt {1 + x} } \right)\sqrt {1 + \sqrt {1 + x} {\text{ }}} - 4\sqrt {1 + \sqrt {1 + x} {\text{ }}} + C \cr} $$
