Answer
$$\frac{{\sqrt 3 }}{{12}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\frac{{dx}}{{{{\left( {4 - {x^2}} \right)}^{3/2}}}}} \cr
& {\text{We set}}{\text{, }}x = 2\sin \theta ,{\text{ }}dx = 2\cos \theta d\theta \cr
& {\text{The new limits on }}\theta {\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}\theta = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6} \cr
& \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}\theta = {\sin ^{ - 1}}\left( {\frac{0}{2}} \right) = 0 \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int_0^1 {\frac{{dx}}{{{{\left( {4 - {x^2}} \right)}^{3/2}}}}} = \int_0^{\pi /6} {\frac{{2\cos \theta d\theta }}{{{{\left( {4 - 4{{\sin }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int_0^{\pi /6} {\frac{{2\cos \theta d\theta }}{{8{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \cr
& {\text{Use the fundamental identity }}{\sin ^2}\theta + {\cos ^2}\theta = 1 \cr
& = \frac{1}{4}\int_0^{\pi /6} {\frac{{\cos \theta d\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \cr
& = \frac{1}{4}\int_0^{\pi /6} {\frac{{\cos \theta d\theta }}{{{{\cos }^3}\theta }}} \cr
& = \frac{1}{4}\int_0^{\pi /6} {{{\sec }^2}\theta d\theta } \cr
& {\text{integrating}} \cr
& = \frac{1}{4}\left( {\tan \theta } \right)_0^{\pi /6} \cr
& = \frac{1}{4}\tan \left( {\frac{\pi }{6}} \right) - \frac{1}{4}\tan \left( 0 \right) \cr
& = \frac{1}{4}\left( {\frac{{\sqrt 3 }}{3}} \right) - \frac{1}{2}\left( 0 \right) \cr
& = \frac{{\sqrt 3 }}{{12}} \cr} $$
