Answer
$$\frac{x}{{\sqrt {1 - {x^2}} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}} \cr
& {\text{substitute }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& = \int {\frac{{\cos \theta d\theta }}{{{{\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)}^{3/2}}}}} \cr
& {\text{simplify}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{3/2}}}}} \cr
& {\text{identity co}}{{\text{s}}^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{\cos \theta d\theta }}{{{{\left( {{{\cos }^2}\theta } \right)}^{3/2}}}}} \cr
& = \int {\frac{{\cos \theta d\theta }}{{{{\cos }^3}\theta }}} \cr
& = \int {{{\sec }^2}\theta } d\theta \cr
& {\text{find antiderivative}} \cr
& = \tan \theta + C \cr
& {\text{write in terms of }}x \cr
& = \frac{x}{{\sqrt {1 - {x^2}} }} + C \cr} $$
