The antiderivative of the integrand is given by $$-\log ^{n+1}(x) ((k+1) \log (x))^{-n-1} \Gamma (n+1,k \log (x)+\log (x))$$ which can be simplified to $$-\log ^{n+1}(x) \text { } E_{-n}((k+1) \log (x))$$ where appears the elliptic integral. As already said by other prticipants the integral to infinity diverges if the lower bound is $0$; if the lower bound is $1$, you get Lucian's result if $\Re(n)>-1$ and $\Re(k)>-1$.
If the lower bound is $a$, the integral is given by $$\frac{\log ^n(a) ((k+1) \log (a))^{-n} (\Gamma (n+1,(k+1) \log (a))-n \Gamma
(n))+\left(\frac{1}{k}+1\right)^{-n} k^{-n} \Gamma (n+1)}{k+1}$$ provided $\Re(n)>-1$ and $\Re(k)>0$.