Write a value

Let, $\log x=t$

Differentiating both sides with respect to $x$

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

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

$y=\int \frac{1}{t^{n}} d t$

Use formula $\int \frac{1}{t^{n}} d t=\frac{t^{-n+1}}{-n+1}$

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

Again, put $t=\log x$

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