<s>[INST] 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
 [/INST] 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
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