Answer
$$ \sqrt {\ln 2} $$
Work Step by Step
$$\eqalign{
& \int_2^{16} {\frac{{dx}}{{2x\sqrt {\ln x} }}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 16,{\text{ }}u = \ln \left( {16} \right) \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ }}u = \ln \left( 2 \right) \cr
& {\text{Write the integral in terms of }}u \cr
& \int_2^{16} {\frac{{dx}}{{2x\sqrt {\ln x} }}} = \frac{1}{2}\int_{\ln 2}^{\ln 16} {\frac{1}{{\sqrt u }}du} \cr
& {\text{Integrate }} \cr
& = \left( {\sqrt u } \right)_{\ln 2}^{\ln 16} \cr
& {\text{Use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \sqrt {\ln 16} - \sqrt {\ln 2} \cr
& = \sqrt {\ln 2^4} - \sqrt {\ln 2} \cr
& = \sqrt {4\ln 2} - \sqrt {\ln 2} \cr
& = 2\sqrt {\ln 2} - \sqrt {\ln 2} \cr
& = \sqrt {\ln 2} \cr } $$
