Answer
$$ - \ln \left| {\frac{{1 + \sqrt {1 - {x^2}} }}{x}} \right| + \sqrt {1 - {x^2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 - {x^2}} }}{x}} dx \cr
& {\text{substitute }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& = \int {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \left( {\cos \theta } \right)d\theta \cr
& = \int {\frac{{\sqrt {{{\cos }^2}\theta } \cos \theta }}{{\sin \theta }}} d\theta \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{\sin \theta }}} d\theta \cr
& {\text{use }}{\cos ^2}\theta = 1 - {\sin ^2}\theta \cr
& = \int {\frac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}} d\theta \cr
& = \int {\left( {\frac{1}{{\sin \theta }} - \frac{{{{\sin }^2}\theta }}{{\sin \theta }}} \right)} d\theta \cr
& = \int {\left( {\csc \theta - \sin \theta } \right)} d\theta \cr
& {\text{find the antiderivative}} \cr
& = - \ln \left| {\csc \theta + \cot \theta } \right| + \cos \theta + C \cr
& {\text{write in terms of }}x \cr
& = - \ln \left| {\frac{1}{x} + \frac{{\sqrt {1 - {x^2}} }}{x}} \right| + \sqrt {1 - {x^2}} + C \cr
& = - \ln \left| {\frac{{1 + \sqrt {1 - {x^2}} }}{x}} \right| + \sqrt {1 - {x^2}} + C \cr} $$
