Answer
$$ - \frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{{3{x^3}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx \cr
& {\text{Refer to the triangle below}} \cr
& \sin \theta = x,{\text{ }}dx = \cos \theta d\theta \cr
& \int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx = \int {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{{{\left( {\sin \theta } \right)}^4}}}} \cos \theta d\theta \cr
& = \int {\frac{{\sqrt {{{\cos }^2}\theta } }}{{{{\sin }^4}\theta }}} \cos \theta d\theta \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{{{\sin }^4}\theta }}} d\theta \cr
& = \int {{{\cot }^2}\theta {{\csc }^2}\theta } d\theta \cr
& = - \int {{{\cot }^2}\theta \left( { - {{\csc }^2}\theta } \right)} d\theta \cr
& {\text{Integrate, recall that }}\frac{d}{{dx}}\left[ {\cot x} \right] = - {\csc ^2}\theta \cr
& = - \frac{1}{3}{\cot ^3}\theta + C \cr
& {\text{Refer to the triangle, }}\cot \theta = \frac{{\sqrt {1 - {x^2}} }}{x} \cr
& = - \frac{1}{3}{\left( {\frac{{\sqrt {1 - {x^2}} }}{x}} \right)^3} + C \cr
& = - \frac{{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}}}{{3{x^3}}} + C \cr
& = - \frac{{{{\left( {1 - {x^2}} \right)}^{3/2}}}}{{3{x^3}}} + C \cr} $$
