Answer
$$\ln \left| {\frac{{1 - \sqrt {1 - {{\ln }^2}x} }}{{\ln x}}} \right| + \sqrt {1 - {{\ln }^2}x} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}{{x\ln x}}} dx \cr
& {\text{set }}y = \ln x,\,\,\,\,dy = \frac{1}{x}dx \cr
& \int {\frac{{\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}{{x\ln x}}} dx = \int {\frac{{\sqrt {1 - {y^2}} }}{y}} dy \cr
& {\text{Using Trigonometric Substitutions}} \cr
& {\text{We set}}{\text{, }}y = \sin \theta,{\text{ }}dy = \cos \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int {\frac{{\sqrt {1 - {y^2}} }}{y}} dy = \int {\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{\sin \theta }}} \left( {\cos \theta } \right)d\theta \cr
& {\text{Use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{\sqrt {{{\cos }^2}\theta } }}{{\sin \theta }}} \left( {\cos \theta } \right)d\theta \cr
& {\text{simplifying, we get:}} \cr
& = \int {\frac{{{{\cos }^2}\theta }}{{\sin \theta }}} d\theta \cr
& = \int {\frac{{1 - {{\sin }^2}\theta }}{{\sin \theta }}} d\theta \cr
& = \int {\left( {\csc \theta - \sin \theta } \right)} d\theta \cr
& {\text{Integrate}} \cr
& = \ln \left| {\csc \theta - \cot \theta } \right| + \cos \theta + C \cr
& {\text{write in terms of }}y,{\text{ }}\csc \theta = \frac{1}{y},\,\,\,\,\,\cot \theta = \frac{{\sqrt {1 - {y^2}} }}{y}{\text{ and cos}}\theta = \sqrt {1 - {y^2}} \cr
& = \ln \left| {\frac{1}{y} - \frac{{\sqrt {1 - {y^2}} }}{y}} \right| + \sqrt {1 - {y^2}} + C \cr
& = \ln \left| {\frac{{1 - \sqrt {1 - {y^2}} }}{y}} \right| + \sqrt {1 - {y^2}} + C \cr
& {\text{write in terms of }}x;{\text{ replace }}y = \ln x \cr
& = \ln \left| {\frac{{1 - \sqrt {1 - {{\ln }^2}x} }}{{\ln x}}} \right| + \sqrt {1 - {{\ln }^2}x} + C \cr} $$
