Answer
$$\ln \left( {2 + \sqrt 3 } \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt 3 } {\frac{{dt}}{{\sqrt {{t^2} + 1} }}} \cr
& {\text{find the antiderivarive using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& \int {\frac{{du}}{{\sqrt {{a^2} + {u^2}} }}} = {\sinh ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \int {\frac{{du}}{{\sqrt {{a^2} + {u^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& \cr
& \int_0^{\sqrt 3 } {\frac{{dt}}{{\sqrt {{t^2} + 1} }}} = \left. {\left( {\ln \left( {t + \sqrt {{t^2} + 1} } \right)} \right)} \right|_0^{\sqrt 3 } \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = \ln \left( {\sqrt 3 + \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 1} } \right) - \ln \left( {0 + \sqrt {{{\left( 0 \right)}^2} + 1} } \right) \cr
& {\text{simplifiyng}} \cr
& = \ln \left( {\sqrt 3 + \sqrt 4 } \right) - \ln \left( {\sqrt 1 } \right) \cr
& = \ln \left( {2 + \sqrt 3 } \right) \cr} $$
