Answer
$$ - \frac{1}{3}{\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)^3} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {1 - {x^2}} \right)}^{1/2}}}}{{{x^4}}}} dx \cr
& {\text{Use trigonometric substitutions:}} \cr
& {\text{ }}x = \sin \theta,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{{{\left( {1 - {x^2}} \right)}^{1/2}}}}{{{x^4}}}} dx = \int {\frac{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{1/2}}}}{{{{\sin }^4}\theta }}} \left( {\cos \theta d\theta } \right) \cr
& {\text{Use the fundamental identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{{{\left( {{{\cos }^2}\theta } \right)}^{1/2}}\cos \theta }}{{{{\sin }^4}\theta }}} d\theta \cr
& {\text{simplifying, we get:}} \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{{{\sin }^4}\theta }}} d\theta \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \left( {\frac{1}{{{{\sin }^2}\theta }}} \right)d\theta \cr
& = \int {{{\cot }^2}\theta } {\csc ^2}\theta d\theta \cr
& {\text{integrate}} \cr
& = - \frac{1}{3}{\cot ^3}\theta + C \cr
& {\text{write in terms of }}x,{\text{ }}\cot \theta = \frac{{\sqrt {1 - {x^2}} }}{x} \cr
& = - \frac{1}{3}{\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)^3} + C \cr} $$
