Answer
\[ = \frac{{{{\left( {\sqrt x + 1\,} \right)}^5}}}{5} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{\,{{\left( {\sqrt x + 1} \right)}^4}}}{{2\sqrt x }}} \,dx \hfill \\
\hfill \\
set\,\,the\,\,substitution \hfill \\
\hfill \\
u = \sqrt x + 1\,\,\,\,\,\,then\,\,\,\,\,du = \frac{1}{{2\sqrt x }}dx \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{\,{{\left( {\sqrt x + 1} \right)}^4}}}{{2\sqrt x }}} \,dx = \int_{}^{} {\,{{\left( {\sqrt x + 1} \right)}^4} \cdot \frac{1}{{2\sqrt x }}\,dx} \hfill \\
\hfill \\
\int_{}^{} {{u^4}du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
\frac{{{u^5}}}{5} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,u = \sqrt x + 1 \hfill \\
\hfill \\
= \frac{{{{\left( {\sqrt x + 1\,} \right)}^5}}}{5} + C \hfill \\
\end{gathered} \]
