Answer
$$\ln 3 + \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {\frac{{dx}}{{{{\left( {1 - {x^2}} \right)}^2}}}} \cr
& {\text{substitute }}x = \sin \theta ,{\text{ }}\,\,\,dx = \cos \theta d\theta \,\,\,\,\theta = {\sin ^{ - 1}}x \cr
& \,\,\,x = 1/2 \to \theta = {\sin ^{ - 1}}\left( {1/2} \right) = \pi /6 \cr
& \,\,\,x = 0 \to \theta = {\sin ^{ - 1}}\left( 0 \right) = 0 \cr
& \cr
& {\text{write the integral in terms of }}\theta \cr
& \int_0^{1/2} {\frac{{dx}}{{{{\left( {1 - {x^2}} \right)}^2}}}} = \int_0^{\pi /6} {\frac{{\cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}d\theta } \cr
& \cr
& {\text{use the pythagorean identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int_0^{\pi /6} {\frac{{\cos \theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^2}}}d\theta } \cr
& = \int_0^{\pi /6} {\frac{{\cos \theta }}{{{{\cos }^4}\theta }}d\theta } \cr
& = \int_0^{\pi /6} {{{\sec }^3}\theta d\theta } \cr
& \cr
& {\text{Where }}\int {{{\sec }^3}\theta } d\theta = \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + \frac{1}{2}\sec \theta \tan \theta + C{\text{ }} \cr
& = \left[ {\frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + \frac{1}{2}\sec \theta \tan \theta } \right]_0^{\pi /6} \cr
& \cr
& {\text{evaluate the limits}} \cr
& = \frac{1}{2}\left[ {\ln \left| {\sec \left( {\frac{\pi }{6}} \right) + \tan \left( {\frac{\pi }{6}} \right)} \right| + \sec \left( {\frac{\pi }{6}} \right)\tan \left( {\frac{\pi }{6}} \right)} \right] - \frac{1}{2}\left[ {\ln \left| {1 + 0} \right| + \sec \left( 0 \right)\tan \left( 0 \right)} \right] \cr
& {\text{Simplify}} \cr
& = \frac{1}{2}\left[ {\ln \left| {\frac{{2\sqrt 3 }}{3} + \frac{{\sqrt 3 }}{3}} \right| + \left( {\frac{{2\sqrt 3 }}{3}} \right)\left( {\frac{{\sqrt 3 }}{3}} \right)} \right] - \frac{1}{2}\left[ 0 \right] \cr
& = \frac{1}{2}\ln \left| {\sqrt 3 } \right| + \frac{1}{3} \cr
& = \ln 3 + \frac{1}{3} \cr} $$
