Answer
$$\frac{\pi }{{12\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& \int_{\sqrt 3 }^3 {\frac{{dt}}{{3 + {t^2}}}} \cr
& = \int_{\sqrt 3 }^3 {\frac{{dt}}{{{{\left( {\sqrt 3 } \right)}^2} + {t^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = \sqrt 3 \cr
& = \left( {\frac{1}{{\sqrt 3 }}ta{n^{ - 1}}\left( {\frac{t}{{\sqrt 3 }}} \right)} \right)_{\sqrt 3 }^3 \cr
& = \frac{1}{{\sqrt 3 }}\left( {ta{n^{ - 1}}\left( {\frac{3}{{\sqrt 3 }}} \right) - ta{n^{ - 1}}\left( {\frac{{\sqrt 3 }}{{\sqrt 3 }}} \right)} \right) \cr
& {\text{simplifying, we get:}} \cr
& = \frac{1}{{\sqrt 3 }}\left( {\frac{\pi }{3} - \frac{\pi }{4}} \right) \cr
& = \frac{1}{{\sqrt 3 }}\left( {\frac{\pi }{{12}}} \right) \cr
& = \frac{\pi }{{12\sqrt 3 }} \cr} $$
