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