Answer
$$\int \frac{6dr}{\sqrt{4-(r+1)^2}}=6\sin^{-1}\Big(\frac{r+1}{2}\Big)+C$$
Work Step by Step
$$A=\int \frac{6dr}{\sqrt{4-(r+1)^2}}$$ $$A=6\int\frac{dr}{\sqrt{4-(r+1)^2}}$$
We set $a=r+1$, which means $$da=dr$$
Therefore, $$A=6\int\frac{da}{\sqrt{4-a^2}}$$
We have $$\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\Big(\frac{x}{a}\Big)+C$$
So, $$A=6\sin^{-1}\Big(\frac{a}{2}\Big)+C$$ $$A=6\sin^{-1}\Big(\frac{r+1}{2}\Big)+C$$
