If $n \in \mathbb{N}$ and $I_n = \int (\log x)^n \, dx$, then $I_n + nI_{n-1} =$
$\frac{(\log x)^{n+1}}{n+1}$
$x(\log x)^n + c$
$\frac{(\log x)^{n-1}}{n}$
$\frac{(\log x)^n}{x}$
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