Answer
$$\int^{3}_{\sqrt3}\frac{dt}{3+t^2}dt=\frac{\pi\sqrt3}{36}$$
Work Step by Step
$$A=\int^{3}_{\sqrt3}\frac{dt}{3+t^2}dt=\int^{3}_{\sqrt3}\frac{dt}{(\sqrt3)^2+t^2}dt$$
We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$
Therefore, $$A=\frac{1}{\sqrt3}\tan^{-1}\Big(\frac{t}{\sqrt3}\Big)\Big]^{3}_{\sqrt3}$$ $$A=\frac{\sqrt3}{3}\Big(\tan^{-1}\sqrt3-\tan^{-1}1\Big)$$ $$A=\frac{\sqrt3}{3}\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$$ $$A=\frac{\sqrt3}{3}\times\frac{\pi}{12}=\frac{\pi\sqrt3}{36}$$
