Inequality 0<x<1: Is $\frac{x-1}{x}<ln(x)<x-1$ True?

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is

\frac{x-1}{x}<ln(x)<x-1

valid for 0<x<1

I think it is I just want to get a second opinion.
 
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One way to look at these things is to examine the functions:
<br /> f(x)=\frac{x-1}{x}-\ln x\quad g(x)=\ln x-(x-1)<br />
and compute the derivatives f'(x) and g'(x) and examine if f(x) and g(x) are increasing or decreasing functions and examine the values for certain points and the inequality should drop out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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