Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt e } {\frac{{dx}}{{x\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}} \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = \sqrt e ,{\text{ }}u = \ln \left( {\sqrt e } \right) = \frac{1}{2} \cr
& {\text{if }}x = 1,{\text{ }}u = \ln \left( 1 \right) = 0 \cr
& {\text{so}} \cr
& \int_1^{\sqrt e } {\frac{{dx}}{{x\sqrt {1 - {{\left( {\ln x} \right)}^2}} }}} = \int_0^{1/2} {\frac{1}{{\sqrt {1 - {u^2}} }}} du \cr
& {\text{find the antiderivative}} \cr
& = \left. {\left( {{{\sin }^{ - 1}}u} \right)} \right|_0^{1/2} \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right) - {\sin ^{ - 1}}\left( 0 \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{6} - 0 \cr
& = \frac{\pi }{6} \cr} $$
