<s>[INST] 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. 
 [/INST] Your integral is equivalent to
$$ \frac{1}{2} \underbrace{\int \frac{2x+1}{\sqrt{x^2 + x + 1}} \mathrm dx}_{=:I_1} - \frac{1}{2} \underbrace{\int \frac{\mathrm d x}{\sqrt{x^2 + x + 1}}}_{=: I_2}.$$
Note that $I_1$ can be easily solved with the substitution $u = x^2 + x + 1$.
For $I_2$ write
$$I_2 = \int \frac{ \mathrm d x}{\sqrt{(x + \frac{1}{2})^2 + \frac{3}{4}}}.$$
Now the substitution $u = \frac{2x+1}{\sqrt{3}}$ will lead you to the integral 
$$\int \frac{\mathrm d u}{\sqrt{u^2 + 1}},$$
which is a standard integral - an antiderivative is $\ln (\sqrt{u^2 + 1} + u)$.
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