Answer
\[\frac{1}{3}\,{\left( {{x^2} - 6x} \right)^{3/2}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {\sqrt {{x^2} - 6x} } \right)\,\left( {x - 3} \right)dx} \hfill \\
Let\,\,u = {x^2} - 6x\,\,,\,\,So\,\,that \hfill \\
du = \,\left( {2x - 6} \right)dx \hfill \\
du = 2\,\left( {x - 3} \right)dx \hfill \\
Then \hfill \\
\frac{1}{2}\int_{}^{} {\,\left( {\sqrt {{x^2} - 6x} } \right)\,\left( {x - 3} \right)dx} = \frac{1}{2}\int_{}^{} {\sqrt u du} \hfill \\
\frac{1}{2}\int_{}^{} {{u^{1/2}}du} \hfill \\
Use\,\,power\,\,rule \hfill \\
\frac{1}{2}\,\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \hfill \\
\frac{1}{3}{u^{3/2}} + C \hfill \\
Substituting\,\,u = {x^2} - 6x\,\,,\,\,gives \hfill \\
\frac{1}{3}\,{\left( {{x^2} - 6x} \right)^{3/2}} + C \hfill \\
\hfill \\
\end{gathered} \]
