Using step-by-step to calculate an integral, I get the following answer

Which is equivalent for restricted $x$ values. Why is Mathematica making this change?

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# Tag: log1x

## Log(x-1) replaced by Log(1-x)

## lower bound for $log(1-x)$ for 0≤$x$≤1-$e^{-1}$

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Using step-by-step to calculate an integral, I get the following answer

Which is equivalent for restricted $x$ values. Why is Mathematica making this change?

How to show $log(1-x) geq -frac{e}{e-1}x$ for 0≤$x$≤1-$e^{-1}$?

I thought about $log(1+x)geq frac{x}{1+x}$, but don’t know how to apply it. Thanks!

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