Answer
$$\ln \left( {\sqrt 2 + 1} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^e {\frac{{dy}}{{y\sqrt {1 + {{\left( {\ln y} \right)}^2}} }}} \cr
& \cr
& {\text{We set}}{\text{, }}\ln y = \tan \theta,{\text{ }}\frac{1}{y}dy = {\sec ^2}\theta d\theta \cr
& \theta = {\tan ^{ - 1}}\left( {\ln y} \right) \cr
& {\text{The new limits on }}\theta {\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}y = e,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {\ln e} \right) = \pi /4 \cr
& \,\,\,\,\,\,{\text{If }}y = 1,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {\ln 1} \right) = 0 \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& \int_1^e {\frac{{dy}}{{y\sqrt {1 + {{\left( {\ln y} \right)}^2}} }}} = \int_0^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}} \cr
& {\text{use the fundamental identity }}{\tan ^2}x + 1 = {\sec ^2}x \cr
& = \int_0^{\pi /4} {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \cr
& = \int_0^{\pi /4} {\sec \theta d\theta } \cr
& {\text{integrating}} \cr
& = \left( {\ln \left| {\sec \theta + \tan \theta } \right|} \right)_0^{\pi /4} \cr
& = \left( {\ln \left| {\sec \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right)} \right|} \right) - \left( {\ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right|} \right) \cr
& {\text{simplifying}} \cr
& = \left( {\ln \left| {\sqrt 2 + 1} \right|} \right) - \left( {\ln \left| {1 + 0} \right|} \right) \cr
& = \ln \left( {\sqrt 2 + 1} \right) \cr} $$
