Answer
$$\frac{4}{{15}}{\left( {1 + \sqrt {1 + x} } \right)^{3/2}}\left( {3\sqrt {1 + x} - 2} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt {1 + \sqrt {1 + x} } } dx \cr
& {\text{Integrate by the substitution method}}{\text{, }} \cr
& \,\,\,\,x + 1 = {u^2},\,\,\,\,\,dx = 2udu \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\sqrt {1 + \sqrt {1 + x} } } dx = \int {\sqrt {1 + \sqrt {{u^2}} } } \left( {2udu} \right) \cr
& = \int {2u\sqrt {1 + u} } du \cr
& \cr
& {\text{Now }}\,\,\,\,{\text{Let }}\,\,\,\,t = 1 + u,\,\,\,u = t - 1,\,\,\,du = dt \cr
& = \int {2\left( {t - 1} \right)\sqrt t } dt \cr
& = 2\int {\left( {t - 1} \right){t^{1/2}}} dt \cr
& = 2\int {\left( {{t^{3/2}} - {t^{1/2}}} \right)} dt \cr
& {\text{Integrate by the power rule}} \cr
& = 2\left( {\frac{{{t^{5/2}}}}{{5/2}} - \frac{{{t^{3/2}}}}{{3/2}}} \right) + C \cr
& = 2\left( {\frac{2}{5}{t^{5/2}} - \frac{2}{3}{t^{3/2}}} \right) + C \cr
& = 4{t^{3/2}}\left( {\frac{1}{5}t - \frac{1}{3}} \right) + C \cr
& \cr
& {\text{Write the integrand in terms of }}u,{\text{ substitute }}1 + u{\text{ for }}t \cr
& = 4{\left( {1 + u} \right)^{3/2}}\left( {\frac{1}{5}\left( {1 + u} \right) - \frac{1}{3}} \right) + C \cr
& = 4{\left( {1 + u} \right)^{3/2}}\left( {\frac{1}{5}u - \frac{2}{{15}}} \right) + C \cr
& = \frac{4}{{15}}{\left( {1 + u} \right)^{3/2}}\left( {3u - 2} \right) + C \cr
& {\text{Write the integrand in terms of }}x,{\text{ substitute }}\sqrt {1 + x} {\text{ for }}u \cr
& = \frac{4}{{15}}{\left( {1 + \sqrt {1 + x} } \right)^{3/2}}\left( {3\sqrt {1 + x} - 2} \right) + C \cr} $$
