Answer
$$\frac{\pi }{{3\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\frac{{dt}}{{4 + 3{t^2}}}} \cr
& = \int_{ - 2}^2 {\frac{{dt}}{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 3 t} \right)}^2}}}} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \sqrt 3 t,{\text{ so that }}du = \sqrt 3 dt \cr
& {\text{the new limits on }}t{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 2,{\text{ then }}u = \sqrt 3 \left( 2 \right) = 2\sqrt 3 \cr
& \,\,\,\,\,\,{\text{If }}t = - 2,{\text{ then }}u = \sqrt 3 \left( 0 \right) = - 2\sqrt 3 \cr
& {\text{then}} \cr
& \int_{ - 2}^2 {\frac{{dt}}{{{{\left( 2 \right)}^2} + {{\left( {\sqrt 3 t} \right)}^2}}}} = \int_{ - 2\sqrt 3 }^{2\sqrt 3 } {\frac{{du/\sqrt 3 }}{{{{\left( 2 \right)}^2} + {u^2}}}} \cr
& = \frac{1}{{\sqrt 3 }}\int_{ - 2\sqrt 3 }^{2\sqrt 3 } {\frac{{du}}{{{{\left( 2 \right)}^2} + {u^2}}}} \cr
& {\text{integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 2 \cr
& = \frac{1}{{\sqrt 3 }}\left( {\frac{1}{2}ta{n^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_{ - 2\sqrt 3 }^{2\sqrt 3 } \cr
& = \frac{1}{{2\sqrt 3 }}\left( {ta{n^{ - 1}}\left( {\frac{u}{2}} \right)} \right)_{ - 2\sqrt 3 }^{2\sqrt 3 } \cr
& = \frac{1}{{2\sqrt 3 }}\left( {ta{n^{ - 1}}\left( {\frac{{2\sqrt 3 }}{2}} \right) - ta{n^{ - 1}}\left( {\frac{{ - 2\sqrt 3 }}{2}} \right)} \right) \cr
& = \frac{1}{{2\sqrt 3 }}\left( {\frac{\pi }{3} + \frac{\pi }{3}} \right) \cr
& = \frac{1}{{2\sqrt 3 }}\left( {\frac{{2\pi }}{3}} \right) \cr
& = \frac{\pi }{{3\sqrt 3 }} \cr} $$
