If $I_n=$$\int \frac{e^{ax}}{x^n}dx$ then $I_n-\frac{a}{n.1}.I_{n-1}=$

$I_n=$$\int e^{ax} .x^{-n}dx$

$=e^{ax}.\frac{x^{-n+1}}{-n+1}-$$\int a.e^{ax} .\frac{x^{-n+1}}{-n+1}dx$

$=\frac{e^{ax}}{\left(1-n\right)x^{n-1}}+\frac{a}{n-1}$$\int \frac{e^{ax}}{x^{n-1}}dx$

$=-\frac{e^{ax}}{\left(n-1\right)x^{n-1}}+\frac{a}{n-1}{I}_{n-1}$

$I_n-\frac{a}{n-1}{I}_{n-1}=\frac{-e^{ax}}{\left(n-1\right)x^{n-1}}$