Answer
$$2\ln \left| {1 + \ln x} \right| - \frac{3}{{1 + \ln x}} + C$$
Work Step by Step
$$\eqalign{
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& \int {\frac{{5 + 2\ln x}}{{x{{\left( {1 + \ln x} \right)}^2}}}dx = } \int {\frac{{5 + 2u}}{{{{\left( {1 + u} \right)}^2}}}du} \cr
& {\text{Decompose }}\frac{{5 + 2u}}{{{{\left( {1 + u} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{5 + 2u}}{{{{\left( {1 + u} \right)}^2}}} = \frac{A}{{1 + u}} + \frac{B}{{{{\left( {1 + u} \right)}^2}}} \cr
& 5 + 2u = A\left( {1 + u} \right) + B \cr
& 5 + 2u = A + Au + B \cr
& 2u + 5 = Au + B + A \cr
& A = 2,{\text{ }}B = 3 \cr
& \frac{{5 + 2u}}{{{{\left( {1 + u} \right)}^2}}} = \frac{2}{{1 + u}} + \frac{3}{{{{\left( {1 + u} \right)}^2}}} \cr
& {\text{Rewrite the integrand}} \cr
& \int {\frac{{5 + 2u}}{{{{\left( {1 + u} \right)}^2}}}du} = \int {\left( {\frac{2}{{1 + u}} + \frac{3}{{{{\left( {1 + u} \right)}^2}}}} \right)du} \cr
& = \int {\frac{2}{{1 + u}}} du + \int {\frac{3}{{{{\left( {1 + u} \right)}^2}}}du} \cr
& {\text{Integrating}} \cr
& = 2\ln \left| {1 + u} \right| - \frac{3}{{1 + u}} + C \cr
& {\text{Write in terms of }}x,{\text{ replace }}u = \ln x \cr
& = 2\ln \left| {1 + \ln x} \right| - \frac{3}{{1 + \ln x}} + C \cr} $$
