Answer
$$\frac{{4{{\left( {1 + \sqrt x } \right)}^{7/2}}}}{7} - \frac{{8{{\left( {1 + \sqrt x } \right)}^{5/2}}}}{5} + \frac{{4{{\left( {1 + \sqrt x } \right)}^{3/2}}}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\sqrt x \cdot \sqrt {1 + \sqrt x } } dx \cr
& {\text{Integrate by the substitution method}}{\text{, }} \cr
& \,\,\,\,x = {u^2},\,\,\,\,\,dx = 2udu \cr
& {\text{Write the integrand in terms of }}u \cr
& \int {\sqrt x \cdot \sqrt {1 + \sqrt x } } dx = \int {\sqrt {{u^2}} \cdot \sqrt {1 + \sqrt {{u^2}} } } \left( {2udu} \right) \cr
& = \int {u\sqrt {1 + u} } \left( {2udu} \right) \cr
& = \int {2{u^2}\sqrt {1 + u} } du \cr
& {\text{Now }}\,\,\,\,{\text{Let }}\,\,\,\,t = 1 + u,\,\,\,u = t - 1,\,\,\,du = dt \cr
& = \int {2{{\left( {t - 1} \right)}^2}\sqrt t } dt \cr
& = 2\int {\left( {{t^2} - 2t + 1} \right){t^{1/2}}} dt \cr
& = 2\int {\left( {{t^{5/2}} - 2{t^{3/2}} + {t^{1/2}}} \right)} dt \cr
& {\text{Integrate by the power rule}} \cr
& = 2\left( {\frac{{{t^{7/2}}}}{{7/2}} - 2\left( {\frac{{{t^{5/2}}}}{{5/2}}} \right) + \frac{{{t^{3/2}}}}{{3/2}}} \right) + C \cr
& = 2\left( {\frac{{2{t^{7/2}}}}{7} - \frac{{4{t^{5/2}}}}{5} + \frac{{2{t^{3/2}}}}{3}} \right) + C \cr
& = \frac{{4{t^{7/2}}}}{7} - \frac{{8{t^{5/2}}}}{5} + \frac{{4{t^{3/2}}}}{3} + C \cr
& \cr
& {\text{Write the integrand in terms of }}u,{\text{ substitute }}1 + u{\text{ for }}t \cr
& = \frac{{4{{\left( {1 + u} \right)}^{7/2}}}}{7} - \frac{{8{{\left( {1 + u} \right)}^{5/2}}}}{5} + \frac{{4{{\left( {1 + u} \right)}^{3/2}}}}{3} + C \cr
& {\text{Write the integrand in terms of }}x,{\text{ substitute }}\sqrt x {\text{ for }}u \cr
& = \frac{{4{{\left( {1 + \sqrt x } \right)}^{7/2}}}}{7} - \frac{{8{{\left( {1 + \sqrt x } \right)}^{5/2}}}}{5} + \frac{{4{{\left( {1 + \sqrt x } \right)}^{3/2}}}}{3} + C \cr} $$
