Answer
$$\frac{3}{2}{\sin ^{ - 1}}\left( {2\left( {r - 1} \right)} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{3dr}}{{\sqrt {1 - 4{{\left( {r - 1} \right)}^2}} }}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = 2\left( {r - 1} \right),{\text{ so that }}du = 2dr,\,\,\,\,\,\,dr = \frac{{du}}{2} \cr
& {\text{then}} \cr
& \int {\frac{{3dr}}{{\sqrt {1 - 4{{\left( {r - 1} \right)}^2}} }}} = 3\int {\frac{{du/2}}{{\sqrt {1 - {u^2}} }}} \cr
& = \frac{3}{2}\int {\frac{{du}}{{\sqrt {1 - {u^2}} }}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = {{\sin }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 1 \cr
& = \frac{3}{2}{\sin ^{ - 1}}\left( u \right) + C \cr
& {\text{write in terms of }}x;{\text{ replace }}2\left( {r - 1} \right){\text{ for }}u \cr
& = \frac{3}{2}{\sin ^{ - 1}}\left( {2\left( {r - 1} \right)} \right) + C \cr} $$
