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!

$x^2-x\ge0\iff x\not\in(0,1)$. Thus, we need to choose whether $x\ge1$ or $x\le0$.

For $x\ge1$, let $x=u+\frac12$ and $u=\frac12\sec(\theta)$. We can assume that $\theta\in[0,\frac\pi2)$ since that gives a full range for $x\ge1$. As is shown in this answer, $\int\sec(\theta)\,\mathrm{d}\theta=\log(\sec(\theta)+\tan(\theta))+C$. Therefore,
$$
\begin{align}
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}
&=\int\frac{\mathrm{d}u}{\sqrt{u^2-\frac14}}\\
&=\int\frac{\sec(\theta)\tan(\theta)\,\mathrm{d}\theta}{\tan(\theta)}\\[9pt]
&=\log(\sec(\theta)+\tan(\theta))+C\\[9pt]
&=\log\left(2x-1+2\sqrt{x^2-x}\right)+C
\end{align}
$$
Since $2x-1+2\sqrt{x^2-x}\ge0$ for all $x\ge1$, absolute values are not needed.

For $x\le0$, we can assume $\theta\in(\frac\pi2,\pi]$. Then, we get
$$
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C
$$
Since $1-2x-2\sqrt{x^2-x}\ge0$ for all $x\le0$, absolute values are not needed.

The absolute values are needed only if we want to try to give a solution for all $x\not\in(0,1)$. However, since the interval of integration cannot span $(0,1)$, we must have either $x\ge1$ or $x\le0$.

Since you have stated that $x\lt0$ in your question, that means we can use the solution for $x\lt0$:
$$
\int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C
$$
and we can solve for $x$. If
$$
u=\log\left(1-2x-2\sqrt{x^2-x}\right)
$$
Then
$$
\begin{align}
e^u&=1-2x-2\sqrt{x^2-x}\\
\left(e^u+2x-1\right)^2&=4x^2-4x\\
e^{2u}+(4x-2)e^u+1&=0\\
x&=\frac12-\frac{e^u+e^{-u}}4\\
&=\frac12(1-\cosh(u))
\end{align}
$$
