If $I_{m,n}=\int \frac{x^m}{{\left(\log x\right)}^n}dxthen(m+1)I_{m,n}-n.I_{m,n+1}=$

$I_{m,n}=\int \frac{x^m}{{\left(\log x\right)}^n}dx$

$=\int {\left(\log x\right)}^{-n}.x^m dx$

$={\left(\log x\right)}^{-n}.\frac{x^{m+1}}{m+1}-$$\int \left(-n\right){\left(\log x\right)}^{-n}.\frac{1}{K}.\frac{x^{m+1}}{m+1}dx$

$I_{m, n}=\frac{x^{m+1}}{\left(m+1\right){\left(\log x\right)}^n}+\frac{n}{m+1}$$\int \frac{x^m}{{\left(\log x\right)}^{n+1}}dx$

$\left(m+1\right)I_{mn}=\frac{x^{m+1}}{{\left(\log x\right)}^n}+n$$\int \frac{x^m}{{\left(\log x\right)}^{n+1}}dx$

$=\frac{x^{m+1}}{{\left(\log x\right)}^n}+n I_{m, n+1}+C$

$\left(m+1\right)I_{mn}-n I_{m, n+1}=\frac{x^{m+1}}{{\left(\log x\right)}^n}+C$