Answer
$$\ln \left( {\frac{2}{{\sqrt 3 }}} \right)$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {\tan x} dx \cr
& {\text{integrate by using the basic integration rule }}\int {\tan x} dx = - \ln \left| {\cos x} \right| + C\,\,\,\left( {{\text{page 694}}} \right) \cr
& then \cr
& = \left( { - \ln \left| {\cos x} \right|} \right)_0^{\pi /6} \cr
& = - \left( {\ln \left| {\cos x} \right|} \right)_0^{\pi /6} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = - \left( {\ln \left| {\cos \frac{\pi }{6}} \right| - \ln \left| {\cos 0} \right|} \right) \cr
& {\text{simplifying}} \cr
& = \ln \left| {\cos 0} \right| - \ln \left| {\cos \frac{\pi }{6}} \right| \cr
& = \ln \left| 1 \right| - \ln \left| {\frac{{\sqrt 3 }}{2}} \right| \cr
& = - \ln \left( {\frac{{\sqrt 3 }}{2}} \right) \cr
& {\text{or}} \cr
& = \ln \left( {\frac{2}{{\sqrt 3 }}} \right) \cr} $$
