Answer
$$\int {{{\ln }^n}x} dx = x{\ln ^n}x - n\int {{{\ln }^{n - 1}}xdx} $$
Work Step by Step
$$\eqalign{
& \int {{{\ln }^n}x} dx \cr
& {\text{substitute }}u = {\ln ^n}x,{\text{ }}du = n{\ln ^{n - 1}}x\left( {\frac{1}{x}} \right)dx \cr
& du = \frac{{n{{\ln }^{n - 1}}x}}{x}dx \cr
& dv = dx,{\text{ }}v = x \cr
& {\text{applying integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {{{\ln }^n}x} dx = \left( {{{\ln }^n}x} \right)\left( x \right) - \int {\left( x \right)\left( {n{{\ln }^{n - 1}}x\left( {\frac{1}{x}} \right)dx} \right)} \cr
& {\text{simplify}} \cr
& \int {{{\ln }^n}x} dx = x{\ln ^n}x - \int {n{{\ln }^{n - 1}}xdx} \cr
& \int {{{\ln }^n}x} dx = x{\ln ^n}x - n\int {{{\ln }^{n - 1}}xdx} \cr} $$
