Answer
$$3\ln \left( {1 + \ln 2} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{3}{{x\left( {1 + \ln x} \right)}}} dx \cr
& = 3\int_1^2 {\frac{1}{{1 + \ln x}}\left( {\frac{1}{x}} \right)} dx \cr
& {\text{use substitution}}{\text{. Let }}u = 1 + \ln x,{\text{ so that }}du = \frac{1}{x}dx \cr
& {\text{the new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = 1 + \ln 2 \cr
& \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 1 + \ln 1 = 1 \cr
& {\text{Then}} \cr
& 3\int_1^2 {\frac{1}{{1 + \ln x}}\left( {\frac{1}{x}} \right)} dx = 3\int_1^{1 + \ln 2} {\frac{1}{u}du} \cr
& {\text{integrate by using }}\int_a^b {\frac{1}{u}} du = \left( {\ln \left| u \right|} \right)_a^b \cr
& = 3\left( {\ln \left| u \right|} \right)_1^{1 + \ln 2} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = 3\left( {\ln \left| {1 + \ln 2} \right| - \ln \left| 1 \right|} \right) \cr
& = 3\ln \left( {1 + \ln 2} \right) \cr} $$
