Integral of $\frac{1}{\sqrt{x^2-x}}dx$ For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was:
$$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$
Which would in my case give:
$$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$
Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$). 
Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!

I'm guessing a bit as to what the OP means, but I think he or she started with the differential equation
$${dx\over dt}=\sqrt{x^2-x}$$
and wishes to find $x$ as a function of $t$.  The formalism
$${dx\over\sqrt{x^2-x}}=dt$$
leads to 
$$\int{dx\over\sqrt{x^2-x}}=\int dt=t+C$$
so that what the OP cites from the table of integrals says
$$\ln\left|2x-1+2\sqrt{x^2-x}\right|=t+C$$
and this is what the OP wants to solve for $x$.  
I'm going to stop at this point until I hear (in a comment) that this is indeed what the OP means.
