Answer
$ - \ln \left| {\frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }}} \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {2 - {x^2}} }}{{{x^2}}}dx} ,{\text{ }}x = \sqrt 2 \sin \theta \cr
& {\text{Let }}x = \sqrt 2 \sin \theta ,{\text{ }}dx = x = \sqrt 2 \cos \theta d\theta \cr
& {\text{Substituting}} \cr
& \int {\frac{{\sqrt {2 - {x^2}} }}{{{x^2}}}dx} = \int {\frac{{\sqrt {2 - {{\left( {\sqrt 2 \cos \theta } \right)}^2}} }}{{{{\left( {\sqrt 2 \cos \theta } \right)}^2}}}\left( {\sqrt 2 \cos \theta } \right)d\theta } \cr
& {\text{Simplify the integrand}} \cr
& = \int {\frac{{\sqrt {2 - 2{{\cos }^2}\theta } }}{{2{{\cos }^2}\theta }}\left( {\sqrt 2 \cos \theta } \right)d\theta } \cr
& = \frac{{\sqrt 2 }}{2}\int {\frac{{\sqrt 2 \sqrt {1 - {{\cos }^2}\theta } }}{{{{\cos }^2}\theta }}\left( {\cos \theta } \right)d\theta } \cr
& = \int {\frac{{\sqrt {1 - {{\cos }^2}\theta } }}{{\cos \theta }}d\theta } \cr
& {\text{Use the identity }}{\cos ^2}\theta + {\sin ^2}\theta = 1 \cr
& = \int {\frac{{\sqrt {{{\sin }^2}\theta } }}{{\cos \theta }}d\theta } \cr
& = \int {\frac{{\sin \theta }}{{\cos \theta }}d\theta } \cr
& {\text{Integrating}} \cr
& = - \ln \left| {\cos \theta } \right| + C \cr
& {\text{Write in terms of }}x,{\text{ use the triangle shown below}} \cr
& \cos \theta = \frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }} \cr
& = - \ln \left| {\frac{{\sqrt {2 - {x^2}} }}{{\sqrt 2 }}} \right| + C \cr} $$
