Answer
$$ - \frac{4}{{\sqrt x }}\left( {\frac{1}{2}\ln x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{\sqrt {{x^3}} }}} dx \cr
& {\text{use the radical property }}\root n \of {{x^m}} = {x^{m/n}} \cr
& = \int {\frac{{\ln x}}{{{x^{3/2}}}}} dx \cr
& = \int {{x^{ - 3/2}}\ln x} dx \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{The integrand has a power of }}u{\text{ multiplying basic functions}} \cr
& {\text{Use formula 50}} \cr
& \left( {50} \right):\,\,\,\,\,\int {{u^n}} \ln udu = \frac{{{u^{n + 1}}}}{{{{\left( {n + 1} \right)}^2}}}\left[ {\left( {n + 1} \right)\ln u - 1} \right] + C \cr
& \int {{x^{ - 3/2}}\ln x} dx \to \,\,\,\,\,u = x,\,\,\,n = 3 \cr
& {\text{By formula 50}} \cr
& \int {{x^{ - 3/2}}\ln x} dx = \frac{{{x^{ - 3/2 + 1}}}}{{{{\left( { - 3/2 + 1} \right)}^2}}}\left[ {\left( { - 3/2 + 1} \right)\ln x - 1} \right] + C \cr
& {\text{simplifying}} \cr
& \int {{x^{ - 3/2}}\ln x} dx = \frac{{{x^{ - 1/2}}}}{{{{\left( { - 1/2} \right)}^2}}}\left[ {\left( { - \frac{1}{2}} \right)\ln x - 1} \right] + C \cr
& \int {{x^{ - 3/2}}\ln x} dx = \frac{{{x^{ - 1/2}}}}{{1/4}}\left[ { - \frac{1}{2}\ln x - 1} \right] + C \cr
& \int {{x^{ - 3/2}}\ln x} dx = \frac{4}{{\sqrt x }}\left( { - \frac{1}{2}\ln x - 1} \right) + C \cr
& \int {{x^{ - 3/2}}\ln x} dx = - \frac{4}{{\sqrt x }}\left( {\frac{1}{2}\ln x + 1} \right) + C \cr} $$
