Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$ Calculate $\int\frac{x\,dx}{\sqrt{x^2 + x + 1}}$.

I tried Euler and hyperbolic substitutions, but both lead to complicated calculations and yet WolframAlpha is able to generate quite simple form of integral.
Hint: $\sqrt{x^2+x+1}=\sqrt{(x+{1\over 2})^2+{3\over 4}}$

$(\sqrt{x^2+1})'={{2x}\over{\sqrt{x^2+1}}}$

Make a variable change.
By completing the square in the denominator, this is equivalent to 

$$\int\frac{x}{\sqrt{(x+\frac12)^2+\frac34}}\,dx.$$

Can you carry on from here?