Answer
$$\frac{{{{\ln }^2}\left( {x - 5} \right)}}{2} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln \left( {x - 5} \right)}}{{x - 5}}} dx \cr
& {\text{integrate by the substitution method}} \cr
& {\text{set }}u = \ln \left( {x - 5} \right){\text{ then }}\frac{{du}}{{dx}} = \frac{1}{{\left( {x - 5} \right)}},\,\,\,\,dx = \left( {x - 5} \right)du \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{\ln \left( {x - 5} \right)}}{{x - 5}}} dx = \int {\frac{u}{{x - 5}}} \left( {x - 5} \right)du \cr
& {\text{cancel common terms}} \cr
& = \int u du \cr
& {\text{integrating}}{\text{,}} \cr
& = \frac{{{u^2}}}{2} + C \cr
& {\text{replace }}\ln \left( {x - 5} \right){\text{ for }}u \cr
& = \frac{{{{\ln }^2}\left( {x - 5} \right)}}{2} + C \cr} $$
