University Calculus: Early Transcendentals (3rd Edition)

$$A=\int x^m(\ln x)^ndx$$ We set $u= (\ln x)^n$ and $dv=x^mdx$ That makes $du=\frac{n(\ln x)^{n-1}}{x}dx$ and $v=\frac{x^{m+1}}{m+1}$ Applying integration by parts $\int udv=uv-\int vdu$, we have $$A=\frac{x^{m+1}}{m+1}(\ln x)^n-\int\frac{x^{m+1}}{m+1}\times\frac{n(\ln x)^{n-1}}{x}dx$$ $$A=\frac{x^{m+1}}{m+1}(\ln x)^n-\frac{n}{m+1}\int x^m(\ln x)^{n-1}dx$$ The reduction formula has been established.