Answer
$$x{\left( {\ln x} \right)^3} - 6x + 6x\ln x - 3x{\left( {\ln x} \right)^2} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\ln x} \right)}^3}dx} \cr
& {\text{Use a table of integrals with forms involving }}\ln u{\text{ functions. }} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& {\text{Formula 91}} \to \int {{{\left( {\ln u} \right)}^n}} du = u{\left( {\ln u} \right)^n} - n\int {{{\left( {\ln u} \right)}^{n - 1}}} + C \cr
& \int {{{\left( {\ln x} \right)}^3}dx} \to n = 3,{\text{ }}u = x \cr
& {\text{Substituting into formula 9}}1 \cr
& \int {{{\left( {\ln x} \right)}^3}} dx = x{\left( {\ln x} \right)^3} - 3\int {{{\left( {\ln x} \right)}^{2 - 1}}} + C \cr
& = x{\left( {\ln x} \right)^3} - 3\int {{{\left( {\ln x} \right)}^2}} + C \cr
& {\text{Use the formula }}\int {{{\left( {\ln u} \right)}^2}du} = u\left[ {2 - 2\ln u + {{\left( {\ln u} \right)}^2}} \right] + C \cr
& = x{\left( {\ln x} \right)^3} - 3x\left[ {2 - 2\ln x + {{\left( {\ln x} \right)}^2}} \right] + C \cr
& = x{\left( {\ln x} \right)^3} - 6x + 6x\ln x - 3x{\left( {\ln x} \right)^2} + C \cr} $$
