Evaluate: $\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}$ 
Evaluate
  $$
  \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} \cdotp
$$

My attempt: 
$$
    I
  = \int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}}
  = \int \frac{\mathrm{d}x}{x\sqrt{\left(x + \frac{1}{2}\right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}
$$
I thought completing the square would bring the integrand into some form, but it did not. Please help.

First use the substitution $x=\frac{1}{t}$ to get
$$\int \frac{\mathrm{d}x}{x\sqrt{x^2+x+1}} =-\int \frac{\mathrm{d}t}{\sqrt{1+t+t^2}}.$$
Now complete the squares and use a tan substitution.
