Simplifying a function of $\ln(x)$ I was asked to convert the function
$$\frac{1}{x \ln x \sqrt{(\ln x)^2-1}}$$
into a function in the expression
$$\frac{1}{x \sqrt{f(x)}}$$
for the domain $x > e$
but I can't seem to find how I can convert it into the answer:
$$\frac{1}{x \ln x \sqrt{(\ln x)^4 - (\ln x)^2}}$$
My instinct was to use the difference of two squares:
$$\frac{1}{x \ln x \sqrt{(\ln x+1)(\ln x-1)}}$$
but I'm still stuck because that method doesn't work. 
Can someone help?
 A: $f(x)=({\log x})^{2}(({\log x})^{2}-1)$ will work.
A: Only a simple calcul
If $x > e$ then
\begin{align}
\frac{1}{x \ln x \sqrt{(\ln x)^2-1}}&=\frac{1}{x \ln x \sqrt{ (\frac{ln x}{ln x})^2(\ln x)^2-1}}\\
&= \frac{1}{x  \dfrac{\ln x}{ln x} \sqrt{ (ln x)^2(\ln x)^2-1}}\\
&=\frac{1}{x \sqrt{ (ln x)^4-(\ln x)^2}} 
\end{align}
A: The function is defined for
\begin{cases}
x>0 \\[4px]
\ln x\ne0 \\[4px]
(\ln x)^2-1>0
\end{cases}
The third condition (together with the first) becomes $\ln x<-1$ or $\ln x>1$, that is, $0<x<1/e$ or $x>e$.
The second condition is $x\ne1$, which is implied by the third condition.
Your function can be rewritten as
$$
\frac{1}{x\ln x\sqrt{(\ln x)^2-1}}=
\begin{cases}
-\dfrac{1}{x\sqrt{(\ln x)^4-(\ln x)^2}} & 0<x<1/e \\[6px]
\dfrac{1}{x\sqrt{(\ln x)^4-(\ln x)^2}} & x>e
\end{cases}
$$
It would be a mistake to just bring $\ln x$ as $(\ln x)^2$ inside the square root for the case $0<x<1/e$, because in this case the value of the function is negative.
The mistake would be the same as writing $1/\sqrt{x^2}=1/x$, which only holds for $x>0$.
