Answer
\[ = 2{\tan ^{ - 1}}\sqrt x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{{x^{\frac{1}{2}}} + {x^{\frac{3}{2}}}}}} \hfill \\
\hfill \\
{\text{factor}}\,\,{\text{the}}\,\,{\text{denominator}} \hfill \\
\hfill \\
= \int_{}^{} {\frac{{dx}}{{{x^{\frac{1}{2}}}\,\left( {1 + x} \right)}}} \hfill \\
\hfill \\
substituting\,\,\,x = {t^2}\,\, \to \,dx\, = 2tdt \hfill \\
\hfill \\
= \int_{}^{} {\frac{{2tdt}}{{t + {t^3}}}} = \,\,\,\,2\int_{}^{} {\frac{{t\,dt}}{{t\,\left( {1 + {t^2}} \right)}}} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= 2\int_{}^{} {\frac{{dt}}{{1 + {t^2}}}} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= 2{\tan ^{ - 1}}t + C \hfill \\
\hfill \\
substituting\,back\,\,t = \,\sqrt x \hfill \\
\hfill \\
= 2{\tan ^{ - 1}}\sqrt x + C \hfill \\
\hfill \\
\end{gathered} \]