$$\int x^{n}ln\ xdx\rfloor n\ne1\begin{cases}u=ln\ x&du=\frac{1}{x}dx\\ \\ \int dv=\int x^{n}dx\Rightarrow v=\frac{x^{n+1}}{n+1}\end{cases}$$
$\frac{n\ln(x)x^{n+1}+\ln(x)x^{n+1}-x^{n+1}}{\left(n+1\right)^{2}}+С$
$n\neq -1$
$\ln(x)x^{n}$
