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

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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