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I'm trying to get a good analytic upper bound for $(\log(x+1))^c$ in terms of $x$ for $x > 1$.

An easy one comes from the fact that $\log(x+1) < 2\log x$, but $2^c (\log x)^c$ seems a bit loose.

By elementary calculus, we have that $(\log(x+1))^c \leqslant (\frac{1}{x} + \log x)^c$, but I can't see how to proceed.

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    $\begingroup$ What form of bound are you looking for? $(\log(x+1))^c$ is pretty simple already... $\endgroup$
    – Igor Rivin
    Dec 9, 2013 at 3:18

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You could take the first Taylor term, or the first three Taylor terms (or if $c<0$, take an even number of Taylor terms).

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