Answer
$$4\sqrt 3 - \frac{{4\pi }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt 3 /2} {\frac{{4{x^2}dx}}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}} \cr
& {\text{We set}}{\text{, }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{The new limits on }}\theta {\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = \sqrt 3 /2,{\text{ then }}\theta = {\sin ^{ - 1}}\left( {\sqrt 3 /2} \right) = \pi /3 \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}\theta = {\sin ^{ - 1}}\left( 0 \right) = 0 \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int_0^{\sqrt 3 /2} {\frac{{4{x^2}dx}}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}} = \int_0^{\pi /3} {\frac{{4{{\left( {\sin \theta } \right)}^2}\cos \theta d\theta }}{{{{\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)}^{3/2}}}}} \cr
& = \int_0^{\pi /3} {\frac{{4{{\sin }^2}\theta \cos \theta d\theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \cr
& {\text{Use the fundamental identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \int_0^{\pi /3} {\frac{{4{{\sin }^2}\theta \cos \theta d\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int_0^{\pi /3} {\frac{{4{{\sin }^2}\theta \cos \theta d\theta }}{{{{\cos }^3}\theta }}} \cr
& = \int_0^{\pi /3} {\frac{{4{{\sin }^2}\theta d\theta }}{{{{\cos }^2}\theta }}} \cr
& = 4\int_0^{\pi /3} {{{\tan }^2}\theta } d\theta \cr
& = 4\int_0^{\pi /3} {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{Integrating, we get:}} \cr
& = 4\left( {\tan \theta - \theta } \right)_0^{\pi /3} \cr
& = 4\left( {\tan \left( {\frac{\pi }{3}} \right) - \frac{\pi }{3}} \right) - 4\left( {\tan \left( 0 \right) - 0} \right) \cr
& = 4\left( {\sqrt 3 - \frac{\pi }{3}} \right) - 4\left( 0 \right) \cr
& = 4\sqrt 3 - \frac{{4\pi }}{3} \cr} $$
