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. 
 A: 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.
A: 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 $\ln (\sqrt{u^2 + 1} + u)$.
A: With $u=x^2+x+1$, the integral would become trivial if you had $x+\frac12$ in the numerator instead of $x$. So write your integral as the difference of two tractable integrals (the second requires $x=\frac{\sqrt{3}}{2}\tan t-\frac12$ instead).
A: 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?
