Write a value

let $\log x=t$

Differentiating on both sides we get,

$\frac{1}{x} d x=d t$

Substituting above equations in $\int \frac{(\log x)^{n}}{x} d x$ we get,

$\int t^{n} d t$

$=\frac{t^{n+1}}{n+1}+c$

$=\frac{(\log x)^{n+1}}{n+1}+c$