Answer
$$\int \frac{5}{9+4r^2}dr=\frac{5}{6}\tan^{-1}\Big(\frac{2r}{3}\Big)+C$$
Work Step by Step
$$A=\int \frac{5}{9+4r^2}dr=5\int\frac{1}{3^2+(2r)^2}dr$$
We set $u=2r$.
Then $$du=2dr$$ $$dr=\frac{1}{2}du$$
Therefore, $$A=\frac{5}{2}\int\frac{du}{3^2+u^2}$$
We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$
So $$A=\frac{5}{2}\times\frac{1}{3}\tan^{-1}\Big(\frac{u}{3}\Big)+C$$ $$A=\frac{5}{6}\tan^{-1}\Big(\frac{u}{3}\Big)+C$$ $$A=\frac{5}{6}\tan^{-1}\Big(\frac{2r}{3}\Big)+C$$
