Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt 3 } {\frac{{3dx}}{{9 + {x^2}}}} \cr
& {\text{integrate using the rule }}\int {\frac{1}{{{a^2} + {u^2}}}du = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C} \cr
& = 3\left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^{\sqrt 3 } \cr
& = \left[ {{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_0^{\sqrt 3 } \cr
& {\text{using the fundamental theorem}} \cr
& = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{3}} \right) - {\tan ^{ - 1}}\left( {\frac{0}{3}} \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{6} - 0 \cr
& = \frac{\pi }{6} \cr} $$
