∫dxx(log⁡x)m is equal to

Here, given integrand is not in any standard form. So,we convert it by substitution method in a standard form and then integrate it. ∫1x(log⁡x)mdx Let  log⁡x=t⇒1x=dtdx⇒dx=xdt∴∫1x(log⁡x)mdx=∫1x(t)mxdt=∫t−mdt=t−m+1−m+1+C=(log⁡x)1−m1−m+C