Answer
$$x{\ln ^2}x - 2x\ln x + 2x + C$$
Work Step by Step
$$\eqalign{
& \int {{{\ln }^2}x} dx \cr
& {\text{a matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{reduction formula 103}} \cr
& \int {{{\ln }^n}xdx} = x{\ln ^n}x - n\int {{{\ln }^{n - 1}}xdx} \cr
& n = 2 \cr
& \int {{{\ln }^2}xdx} = x{\ln ^2}x - 2\int {{{\ln }^{2 - 1}}xdx} \cr
& \int {{{\ln }^2}xdx} = x{\ln ^2}x - 2\int {\ln xdx} \cr
& {\text{formula 11 }}\int {\ln x} dx = x\ln x - x + C \cr
& \int {{{\ln }^2}xdx} = x{\ln ^2}x - 2\left( {x\ln x - x} \right) + C \cr
& \int {{{\ln }^2}xdx} = x{\ln ^2}x - 2x\ln x + 2x + C \cr} $$
