Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help? 

We can eliminate the square root with an Euler substitution of
$$t = \sqrt{x^2-x} - x \implies x = -\frac{t^2}{1+2t} \implies dx = -\frac{2t(1+t)}{(1+2t)^2} \, dt$$
Then
$$\begin{align*}
I &= \int \sqrt{x^2-x} \, dx \\[1ex]
&= -2 \int \left(t-\frac{t^2}{1+2t}\right) \frac{t(1+t)}{(1+2t)^2} \, dt \\[1ex]
&= -2 \int \frac{t^2(1+t)^2}{(1+2t)^3} \, dt \\[1ex]
&= - \frac18 \int \left(1 + 2t - \frac2{1+2t} + \frac1{(1+2t)^3}\right) \, dt
\end{align*}$$
