不定积分：

设 $I_n=\int (\ln x)^n\,dx$（$x>0$）。分部积分，取 $u=(\ln x)^n,\ dv=dx$，则 $du=\frac{n(\ln x)^{n-1}}{x}dx,\ v=x$：

$$I_n=x(\ln x)^n-n\int (\ln x)^{n-1}dx=x(\ln x)^n-nI_{n-1}.$$

递推展开可写成闭式：

$$\boxed{\int (\ln x)^n dx=x\sum_{k=0}^{n}(-1)^k\frac{n!}{(n-k)!}(\ln x)^{n-k}+C}.$$

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英文解析

Let $I_n=\int (\ln x)^n\,dx$ ($x>0$). Using integration by parts with$u=(\ln x)^n$and$dv=dx$, we have$du=\frac{n(\ln x)^{n-1}}{x}dx$and$v=x$:

$$I_n=x(\ln x)^n-n\int (\ln x)^{n-1}dx=x(\ln x)^n-nI_{n-1}.$$

The recursive expansion can be written in closed form as:

$$\boxed{\int (\ln x)^n dx=x\sum_{k=0}^{n}(-1)^k\frac{n!}{(n-k)!}(\ln x)^{n-k}+C}.$$