Integration: $\int\frac{1}{(x^2+x+1)^{1/2}} dx$ Find the value of $$\int\frac{1}{(x^2+x+1)^{1/2}} dx$$ 
Anyone can provide hint on how to integrate this, and how you know what method to use? (I mean, is there any general guideline to follow for solving?)
Thank you!

If you make the Euler substitution $t=x+\sqrt{x^2+x+1}$, then $\displaystyle x=\frac{t^2-1}{2t+1}$ and $\displaystyle dx=\frac{2(t^2+t+1)}{(2t+1)^2}dt$;
so $\displaystyle\int\frac{1}{\sqrt{x^2+x+1}}dx=\int\frac{1}{t-\frac{t^2-1}{2t+1}}\cdot\frac{2(t^2+t+1)}{(2t+1)^2}dt=\int\frac{2}{2t+1}dt=\ln\lvert2t+1\lvert+C$
$\displaystyle=\ln\left(2x+2\sqrt{x^2+x+1}+1\right)+C$
