Answer
\[ = \frac{3}{2}\,{\left( {x - 4} \right)^{\frac{3}{2}}} + 8\,{\left( {x - 4} \right)^{\frac{1}{2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{x}{{\sqrt {x - 4} }}\,dx} \hfill \\
\hfill \\
set \hfill \\
\hfill \\
u = x - 4\,\,\,then\,\,\,s = u + 4 \hfill \\
\hfill \\
and\,\,\,dx = du \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int_{}^{} {\frac{{u + 4}}{{{u^{\frac{1}{2}}}}}du} \hfill \\
\hfill \\
Distribute \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {{u^{\frac{1}{2}}} + 4{u^{\frac{{ - 1}}{2}}}} \right)du} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \frac{3}{2}{u^{\frac{3}{2}}} + 8{u^{\frac{1}{2}}} + C \hfill \\
\hfill \\
replace\,\,u\,\,with\,\,\,u = x + 4 \hfill \\
\hfill \\
= \frac{3}{2}\,{\left( {x - 4} \right)^{\frac{3}{2}}} + 8\,{\left( {x - 4} \right)^{\frac{1}{2}}} + C \hfill \\
\hfill \\
\end{gathered} \]
