## Calculus: Early Transcendentals (2nd Edition)

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