不定积分：

对 $n\ne -1$，分部积分取 $u=\ln x,\ dv=x^n dx$，则 $du=dx/x$，$v=\frac{x^{n+1}}{n+1}$：

$$\int x^n\ln x\,dx=\frac{x^{n+1}}{n+1}\ln x-\frac{1}{n+1}\int x^n dx =\boxed{\frac{x^{n+1}}{n+1}\ln x-\frac{x^{n+1}}{(n+1)^2}+C}.$$

若 $n=-1$，则 $\int \frac{\ln x}{x}dx=\frac12(\ln x)^2+C$。

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英文解析

For $n \ne -1$, using integration by parts with $u=\ln x$and $dv=x^n dx$, we have $du=\frac{dx}{x}$and $v=\frac{x^{n+1}}{n+1}$:

$$\int x^n\ln x\,dx=\frac{x^{n+1}}{n+1}\ln x-\frac{1}{n+1}\int x^n dx =\boxed{\frac{x^{n+1}}{n+1}\ln x-\frac{x^{n+1}}{(n+1)^2}+C}.$$

If $n=-1$, then $\int \frac{\ln x}{x}dx=\frac12(\ln x)^2+C$.