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