Answer
$$\frac{{{{\ln }^4}x}}{4} + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {\ln x} \right)}^3}}}{x}} dx \cr
& {\text{Integrate by substitution method}} \cr
& {\text{Let }}u = \ln x,\,\,\,\,du = \frac{1}{x}dx,\,\,\,\,dx = xdu \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{{{\left( {\ln x} \right)}^3}}}{x}} dx = \int {\frac{{{u^3}}}{x}} xdu \cr
& {\text{Cancel common factor }}x \cr
& = \int {{u^3}} du \cr
& {\text{integrating}} \cr
& = \frac{{{u^4}}}{4} + C \cr
& {\text{replacing }}u = \ln x \cr
& = \frac{{{{\ln }^4}x}}{4} + C \cr} $$
