∫dxx(logx)m is equal to
(logx)mm+C
(logx)m−1m−1+C
(logx)1−m1−m+C
(logx)1−mm+C
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(logx)mdx Let logx=t⇒1x=dtdx⇒dx=xdt∴∫1x(logx)mdx=∫1x(t)mxdt=∫t−mdt=t−m+1−m+1+C=(logx)1−m1−m+C