Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 396: 82

\begin{align*} \int(\ln x)^{k} d x&=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x \end{align*}

Given $$\int(\ln x)^{k} d x$$ Let \begin{align*} u&= (\ln x)^k \ \ \ \ \ \ \ dv=dx\\ du& =\frac{k}{x}(\ln x)^{k-1}\ \ \ \ \ v=x \end{align*} Then \begin{align*} \int(\ln x)^{k} d x&=x(\ln x)^{k}-k \int(\ln x)^{k-1} d x \end{align*}