Answer
\[ = \frac{4}{3}\sqrt {1 + \sqrt x } \,\left( {\sqrt x - 2} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\sqrt {1 + \sqrt x } }}} \hfill \\
\hfill \\
set\,\,\,x = \,{\left( {{u^2} - 1} \right)^2}\,\,\,\,\, \hfill \\
then\,\,\,dx = 4\,\left( {{u^2} - 1} \right)udu \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {1 + \sqrt x } }}} \, = \int_{}^{} {\frac{{4\,\left( {{u^2} - 1} \right)udu}}{u}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= 4\int_{}^{} {\,\left( {{u^2} - 1} \right)du} \hfill \\
\hfill \\
{\text{Integrate}} \hfill \\
\hfill \\
= 4\,\,\left[ {\frac{{{u^3}}}{3} - u} \right] + C \hfill \\
\hfill \\
substitute\,\,for\,\,\,u \hfill \\
\hfill \\
= \frac{4}{3}\sqrt {1 + \sqrt x } \,\left( {\sqrt x - 2} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
