Answer
$$\frac{1}{{2\ln 2}}$$
Work Step by Step
$$\eqalign{
& \int_2^4 {\frac{{dx}}{{x{{\left( {\ln x} \right)}^2}}}} \cr
& {\text{Use substitution:}}\cr
& {\text{Let }}u = \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 4,{\text{ }}u = \ln \left( 4 \right) \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ }}u = \ln \left( 2 \right) \cr
& {\text{write the integral in terms of }}u \cr
& \int_2^4 {\frac{{dx}}{{x{{\left( {\ln x} \right)}^2}}}} = \int_{\ln 2}^{\ln 4} {\frac{1}{{{u^2}}}du} \cr
& = \int_{\ln 2}^{\ln 4} {{u^{ - 2}}du} \cr
& {\text{integrate by using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \left( {\frac{{{u^{ - 1}}}}{{ - 1}}} \right)_{\ln 2}^{\ln 4} \cr
& = \left( {\frac{1}{u}} \right)_{\ln 4}^{\ln 2} \cr
& {\text{use fundamental theorem of calculus: }}\cr
& \int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{1}{{\ln 2}} - \frac{1}{{\ln 4}} \cr
& {\text{simplifying}} \cr
& = \frac{1}{{\ln 2}} - \frac{1}{{\ln {2^2}}} \cr
& = \frac{1}{{\ln 2}} - \frac{1}{{2\ln 2}} \cr
& = \frac{1}{{2\ln 2}} \cr} $$
