Prove a real valued function is increasing Prove $$-\frac{x(1-x^c)\ln x}{(1-x)^2}$$
 $c&gt;0$ is an increasing function on $[0,1]$.

I have a relatively cumbersome proof sketched below. I would like to see other ideas, particularly simpler ones.



Define $$g(x,c) = -\frac{x(1-x^c)\ln x}{c(1-x)^2}.$$ Then 
$$\frac{g(x,c_2)}{g(x,c_1)}=\frac{c_1(1-x^{c_2})}{c_2(1-x^{c_1})},$$
$0\le c_1&lt;c2$ is increasing for $x\in[0,1]$. Thus, if $g(x,c_1)$ increases then $g(x,c_2)$ also increases.

Take $c_1=0$, $g(x,0)=\frac{x(\ln x)^2}{(1-x)^2}$. We can show $g(x,0)$ is increasing in $x$ by substituting $x = e^{-2t}$. Therefore, $g(x,c)$ is increasing in $x$ for $c&gt;0$.
$$-\frac{x(1-x^c)\ln x}{(1-x)^2} \;\; = \;\; x \cdot (x^c - 1) \cdot \ln x \cdot \frac{1}{(1-x)^2}$$

Note that this is a product of $4$ functions, each of which is increasing on the open interval defined by $0 &lt; x &lt; 1.$

NOTE: This is incorrect reasoning, as explained in the comments. I'll leave it up so that others can see why the kind of simplistic approach I used doesn't work, and in case I think of something reasonably simple that does work.
$$-\frac{x(1-x^c)\ln x}{(1-x)^2} \;\; = \;\; x \cdot (x^c - 1) \cdot \ln x \cdot \frac{1}{(1-x)^2}$$

Note that this is a product of $4$ functions, each of which is increasing on the open interval defined by $0 &lt; x &lt; 1.$

NOTE: This is incorrect reasoning, as explained in the comments. I'll leave it up so that others can see why the kind of simplistic approach I used doesn't work, and in case I think of something reasonably simple that does work.