View source for Integral of (z^n)log(z)dz=(z^(n+1)/(n+1))log(z)-z^(n+1)/(n+1)^2 for integer n neq -1

==Theorem==
The following formula holds for integers $n \neq -1$:
$$\displaystyle\int z^n \log(z) \mathrm{d}z= \dfrac{z^{n+1}\log(z)}{n+1} - \dfrac{z^{n+1}}{(n+1)^2}+C,$$
where $\log$ denotes the [[logarithm]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Antiderivative of the logarithm|next=findme}}: $4.1.50$

[[Category:Theorem]]
[[Category:Unproven]]