Answer
$$\frac{\sqrt 3 \pi}{9}.$$
Work Step by Step
Assume that $u=x/\sqrt 3$, then $du= \frac{1}{\sqrt 3}dx$, and when $x:0\to 3$ then $u:0\to \sqrt 3$ and hence we get
$$\int_0^{3} \frac{ dx}{3+x^2}=\int_0^{\sqrt 3} \frac{\sqrt 3 du}{3+3u^2}=\frac{\sqrt 3}{3}\int_0^{\sqrt 3} \frac{ du}{1+u^2}\\ =\frac{\sqrt 3}{3} \tan^{-1}u|_0^{\sqrt 3}=\frac{\sqrt 3}{3}( \tan^{-1}\sqrt{3}- \tan^{-1}0)=\frac{\sqrt 3 \pi}{9}.$$
