Answer
\[ = \ln \left| {x - 3 + \sqrt {\,{{\left( {x - 3} \right)}^2} - 9} } \right| + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 6x} }}} \hfill \\
\hfill \\
set\,\,\,\,\,\,\,x - 3 = u\,\,\,then\,\,\,\,\,\,dx = du \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 6x} }} = \int_{}^{} {\frac{{du}}{{\sqrt {{u^2} - {3^2}} }}} } \hfill \\
\hfill \\
{\text{using}}\,\,the\,\,formula\, \hfill \\
\, \hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - {a^2}} }}} = \ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_{}^{} {\frac{{du}}{{\sqrt {{u^2} - {3^2}} }}} = \ln \left| {u + \sqrt {{u^2} - {3^2}} } \right| + C \hfill \\
\hfill \\
subsitute\,back \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {{x^2} - 6x} }}} = \ln \left| {x - 3 + \sqrt {\,{{\left( {x - 3} \right)}^2} - {3^2}} } \right| + C \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \ln \left| {x - 3 + \sqrt {\,{{\left( {x - 3} \right)}^2} - 9} } \right| + C \hfill \\
\end{gathered} \]