Answer
$$\int \frac{3dr}{\sqrt{1-4(r-1)^2}}=\frac{3}{2}\sin^{-1}\Big(2(r-1)\Big)+C$$
Work Step by Step
$$A=\int \frac{3dr}{\sqrt{1-4(r-1)^2}}$$ $$A=3\int\frac{dr}{\sqrt{1-(2(r-1))^2}}$$
We set $a=2(r-1)$, which means $$da=2dr$$ $$dr=\frac{1}{2}da$$
Therefore, $$A=\frac{3}{2}\int\frac{da}{\sqrt{1-a^2}}$$ $$A=\frac{3}{2}\sin^{-1}a+C$$ $$A=\frac{3}{2}\sin^{-1}\Big(2(r-1)\Big)+C$$
