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? 
 A: hint
$$\sqrt {x^2-x}=\frac {1}{2}\sqrt {(2x-1)^2-1} $$
then put $$2x-1=\cosh (t) $$
and use
$$\cosh^2(t)-1=\sinh^2(t)$$
A: $$\int \sqrt{x^2-x} \ dx$$
$$=\int \sqrt{(x-\frac12)^2-\frac14}dx$$
Apply u-substitution:$u=x-\frac12$
$$=\int\sqrt{u^2-\frac14}du$$
Apply Trig Substitution: $u=\frac12 \sec(t)$
$$=\int\frac{\sec(t)\tan(t)\sqrt{\sec^2(t)-1}}{4}dt$$
$$\frac14\int \sec(t)\tan(t)\sqrt{\sec^2(t)-1}dt$$
Use the identity:$\sec^2(x)=1+\tan^2(x)$
$$\frac14 \int \sqrt{-1+1+\tan^2(t)}(\sec(t)\tan(t))dt$$
$$\frac14 \int \sqrt{\tan^2(t)}(\sec(t)\tan(t))dt$$
$$=\frac14 \int \tan(t)\sec(t)\tan(t)dt$$
$$=\frac14 \int \tan^2(t)\sec(t)dt$$
Using the identity:$\tan^2(x)=-1+\sec^2(x)$
$$=\frac14 \int (-1+\sec^2(t))\sec(t)dt$$
$$\frac14 \int -\sec(t)+\sec^3(t)dt$$
$$\frac14(-\int \sec(t)dt+\int \sec^3(t)dt)$$
Note that $\int \sec(t)dt=\ln |\tan(t)+\sec(t)|$
and $\int \sec^3(t)dt=\frac{\sec^2(t)\sin(t)}{2}+\frac12 \ln |\tan(t)+\sec(t)|$
$$=\frac14(-\ln |\tan(t)+\sec(t)|+\frac{\sec^2(t)\sin(t)}{2}+\frac12 \ln |\tan(t)+\sec(t)|)$$
And finally after substituting back the "u and t", we get
$$\int \sqrt{x^2-x}dx=\frac18 (4x^2 \sqrt{-\frac{1}{(2x-1)^2}+1}-4x \sqrt{-\frac{1}{(2x-1)^2}+1}+ \sqrt{-\frac{1}{(2x-1)^2}+1}-\ln|2x\sqrt{-\frac{1}{(2x-1)^2}+1}-\sqrt{-\frac{1}{(2x-1)^2}+1}+2x-1|)+C$$
A: complete the square
$$\begin{align}
\int\sqrt{x^2-x} \ dx &=I \\
x^2-x&=\left(x-\frac{1}{2}\right)^2-\frac{1}{4} \\
\text{set} \quad u&=x-\frac{1}{2} \quad \text{then} \quad dx=du\\
I&=\int\sqrt{u^2-a} \ du \quad \text{such that} \quad a=\frac{1}{4}\\
\end{align}$$
This type of integral is well known.  You should now use a trig substitution.
A: $$\int\sqrt{x^2-x}\; dx=\int\sqrt{x}\; \sqrt{x-1}\; dx\qquad\qquad x\rightarrow\; \cosh^2\theta\quad dx\rightarrow2\cosh\theta\sinh\theta\; d\theta$$
$$=2\int\cosh^2\theta\sinh^2\theta\; d\theta\ =2\int\cosh^2\theta(\cosh^2\theta-1)\; d\theta\ = 2\int\cosh^4\theta\; d\theta -2\int\cosh^2\theta\; d\theta\ $$
$$=\frac{1}{2}\int(1+\cosh(2\theta))^2\;d\theta\ -\int(1+cosh(2\theta))\; d\theta$$
$$\frac{1}{2}\int(1+2\cosh(2\theta)+\cosh^2(2\theta))\; d\theta\ -\theta-\frac{1}{2}\sinh(2\theta)+C$$
$$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)+C\ +\frac{1}{4}\int(1+\cosh(4\theta))\; d\theta$$
$$\frac{\theta}{2}+\frac{1}{2}\sinh(2\theta)-\theta-\frac{1}{2}\sinh(2\theta)\ +\frac{\theta}{4}+\frac{1}{16}\sinh(4\theta)+C$$
$$\frac{1}{16}\sinh(4\theta)-\frac{\theta}{4}+C$$
$$\theta=\cosh^{-1}(\sqrt{x})$$
$$\frac{1}{16}\sinh(4\cosh^{-1}(\sqrt{x}))-\frac{1}{4}\cosh^{-1}(\sqrt{x})+C$$
This is the short answer if you don't want so much details. Of course if you have enough skills can try using Integration By Parts or another Trig. Subs. 
A: Integrate by parts instead as follows
\begin{align}\int\sqrt{x^2-x}\ dx
= &\int\frac{\sqrt{x^2-x}}{2(x-\frac12)}d[(x-\frac12)^2]\\
= &\ \frac12(x-\frac12) \sqrt{x^2-x}-\frac18\int \frac 1{ \sqrt{x^2-x}}dx \\
 =& \ \frac12(x-\frac12) \sqrt{x^2-x}
-\frac18\tanh^{-1}\frac{\sqrt{x^2-x}}{x-\frac12}
\end{align}
A: 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
$$\begin{align*}
I &= \int \sqrt{x^2-x} \, dx \\[1ex]
&= -2 \int \left(t-\frac{t^2}{1+2t}\right) \frac{t(1+t)}{(1+2t)^2} \, dt \\[1ex]
&= -2 \int \frac{t^2(1+t)^2}{(1+2t)^3} \, dt \\[1ex]
&= - \frac18 \int \left(1 + 2t - \frac2{1+2t} + \frac1{(1+2t)^3}\right) \, dt
\end{align*}$$
