$I_{m, n}=\int\limits_0^1 x^m(\ln x)^n d x=$

$\frac{-n}{m+1} I_{m, n-1}$

$I_{m, n}=\int\limits_0^1 x^m(\log x)^n d x$

(Integrating by parts taking (log x)ⁿ as first function)

$=\left[(\log x)^n . \frac{x^{m+1}}{m+1}\right]_0^1-\int\limits n(\log x)^{n-1} . \frac{1}{x} . \frac{x^{m+1}}{m+1} d x$

$=0-\frac{n}{m+1} \int\limits_0^1 n^m(\log x)^{n-1}=-\frac{n}{m+1} I_{m, n-1}$

Hence (3) is the correct answer.