Prove with induction that:
$$\int x^n(\ln(x))^n\,\mathrm{d}x=\sum_{k=0}^{n}\frac{(-1)^kn!}{(n-k)!(n+1)^{k+1}}x^{n+1}(\ln(x))^{n-k}$$
This is a question of a previous exam. Im trying to do this problem but im stuck on getting back to the induction hypothesis. Can someone help, Integration by parts didn't really work?
This is what I've tried so far:
$$\int x^{n+1}(\ln(x))^{n+1}\mathrm{d}x = (\ln(x))^{n+1}\frac{x^{n+2}}{n+2}-\frac{n+1}{n+2}\int x^{n+1}(\ln(x))^{n}\mathrm{d}x$$ I cant get back to my induction hypothesis.
What you did can be made to work. Let
$$I_x(m,n) = \int x^{m}\ln(x)^{n}\mathrm{d}x$$
Apply parts as you did to get $$I_x(m,n) = \frac{x^{m+1}}{m+1} \ln(x)^n - \frac{n}{m+1} I(m,n-1)$$
We also have $$I_x(m,0) = \frac{x^{m+1}}{m+1}$$
and you can now prove the summation formula by induction using the above relations.