Answer
$$6{\sin ^{ - 1}}\left( {\frac{{r + 1}}{2}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{6dr}}{{\sqrt {4 - {{\left( {r + 1} \right)}^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = r + 1,{\text{ so that }}du = dr \cr
& {\text{then}} \cr
& \int {\frac{{6dr}}{{\sqrt {4 - {{\left( {r + 1} \right)}^2}} }}} = 6\int {\frac{{du}}{{\sqrt {4 - {u^2}} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = 6{\sin ^{ - 1}}\left( {\frac{u}{2}} \right) + C \cr
& {\text{write in terms of }}r;{\text{ replace }}r + 1{\text{ for }}u \cr
& = 6{\sin ^{ - 1}}\left( {\frac{{r + 1}}{2}} \right) + C \cr} $$
