Answer
$$\ln \left( {2 + \sqrt 5 } \right)$$
Work Step by Step
$$\eqalign{
& \int_1^{{e^2}} {\frac{{dx}}{{x\sqrt {{{\ln }^2}x + 1} }}} \cr
& {\text{set }}u = \ln x{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \frac{1}{x}dx, \cr
& {\text{switch the limits of integration}} \cr
& u = \ln x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \to u = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = {e^2} \to u = 2 \cr
& {\text{use the change of variable}} \cr
& \int_1^{{e^2}} {\frac{{dx}}{{x\sqrt {{{\ln }^2}x + 1} }}} = \int_0^2 {\frac{{du}}{{\sqrt {{u^2} + 1} }}} \cr
& {\text{using the Integral formulas in the Theorem 6}}{\text{.12 }}\left( {formula\,\,2} \right)\,\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = {{\sinh }^{ - 1}}\frac{x}{a} + C{\text{.}}} \cr
& {\text{then}} \cr
& \int_0^2 {\frac{{du}}{{\sqrt {{u^2} + 1} }}} = \left( {{{\sinh }^{ - 1}}u} \right)_0^2 \cr
& {\text{evaluate the limits of integration}} \cr
& = {\sinh ^{ - 1}}2 - {\sinh ^{ - 1}}0 \cr
& = {\sinh ^{ - 1}}2 \cr
& {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}: \cr
& {\text{ }}{\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right){\text{ then}}{\text{,}} \cr
& {\sinh ^{ - 1}}2 = \ln \left( {2 + \sqrt {{2^2} + 1} } \right) \cr
& = \ln \left( {2 + \sqrt 5 } \right) \cr} $$