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