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