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!

$$\int \frac{dx}{\sqrt{x^2-x}} = \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}}$$
Setting $ x - \frac{1}{2} = t $
$$ \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}} = \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} $$
It's possible to do this using a trig substitution, but if you want the inverse function, a better way is to use a hyperbolic substitution. Let $t = \frac{1}{2}\cosh u \Rightarrow dt = \frac{1}{2} \sinh u$
$$ \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} = \int \frac{\frac{1}{2}\sinh u \, du}{\sqrt{\frac{1}{4}(\cosh^2 u - 1)}} = \int \frac{\frac{1}{2}\sinh u \, du}{\frac{1}{2}\sinh u} = \int du = u + C $$
Working backwards is easy since you already have $t$ as a function of $u$
$$t = \frac{1}{2}\cosh u \Rightarrow x = \frac{1}{2}\cosh u + \frac{1}{2}$$
Where $u$ is the integral in question
