any other method for evaluating $\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx$? I tried below and its getting tedious :
$\begin{align}\\ \int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx &=  \int\limits \frac{(2x-1)+ x^2-x+2 }{ \sqrt{x^2-x+1} } dx \\~\\  &=  \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} }  +   \int\limits \frac{ (x^2-x+1 )dx}{ \sqrt{x^2-x+1} }  +   \int\limits \frac{ dx }{ \sqrt{x^2-x+1} }  \\~\\&\cdots\\~\\ \end{align}$
wolfram shows very much simplified answer :
http://www.wolframalpha.com/input/?i=%5Cint+%28x%5E2%2Bx%2B1%29%2F%28sqrt%28x%5E2-x%2B1%29%29
I'm wondering if there is any nice way to work this
 A: $$
\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx = \frac{2}{\sqrt{3}}\int \frac{x^2+x+1 }{\sqrt{\frac{4}{3}\left(x-\frac{1}{2}\right)^2 + 1} }
$$
set $t= \frac{2}{\sqrt{3}}\left(x-\frac{1}{2}\right)$
we obtain
$$
\frac{1}{4}\int\frac{3t^2+4\sqrt{3}t + 7}{\sqrt{t^2+1}}
$$
This is the same way as yours but a little cleaner to solve :).
A: I'm adding a separate answer as it is markedly different from my other answer.
In fact, your way is not too tough either 
$$\int\limits \frac{ x^2+x+1 }{ \sqrt{x^2-x+1} } dx =\cdots=  \int\limits \frac{(2x-1)dx}{ \sqrt{x^2-x+1} }  +   \int\limits \sqrt{x^2-x+1}\ dx   +   \int\limits \frac{ dx }{ \sqrt{x^2-x+1} } $$
For the first integral, observe that $\displaystyle\frac{d(x^2-x+1)}{dx}=2x-1$
for the rest two as $\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4,$ set $2x-1=y$
and utilize $\#1,\#8$ of this
A: HINT:
I think using Trigonometric substitution of 
$\displaystyle x^2-x+1=\frac{(2x-1)^2+(\sqrt3)^2}4$ 
with $2x-1=\sqrt3\tan\theta$ won't be a bad way to start with
