An interesting identity involving radicals

This identity came out while solving the indefinite integral

$$I = \int \frac{dx}{(x+1)\sqrt{x^2+x+1}} \tag{1} $$

I got my answer as 

$$I = F(x) = \ln { \frac{-1 + \sqrt{x^2+x+1}}{-1-2x+\sqrt{x^2+x+1}}} \tag{2}$$

but the answer given in the book was 

$$I = G(x) = \ln {\frac{x + \sqrt{x^2+x+1}}{x + 2 + \sqrt{x^2+x+1}}} \tag{3}$$

Checking the two answers with WA shows that both are correct.

So it is natural then to ask... what is the relation between these two expressions?

After some short struggle, I found that the relation is as follows:

$$ \frac{-1 + \sqrt{x^2+x+1}}{-1-2x+\sqrt{x^2+x+1}}= (-1) \cdot \frac{x + \sqrt{x^2+x+1}}{x + 2 + \sqrt{x^2+x+1}} \tag{4}$$

One can easily prove this by letting $a = \sqrt{x^2+x+1}$ and then doing some simple algebraic manipulations.

Of course $(4)$ is true only for those real values of $x$ for which both sides are well-defined. 

The curious thing is that even though $F(x)$ and $G(x)$ have identical derivatives (identical when viewed as an expression of $x$, I mean), they are never simultaneously well-defined. Why? Because when $$f(x) = \frac{-1 + \sqrt{x^2+x+1}}{-1-2x+\sqrt{x^2+x+1}}$$ and $$g(x) = \frac{x + \sqrt{x^2+x+1}}{x+2+\sqrt{x^2+x+1}}$$ are both defined and non-zero, they have opposite signs (as $(4)$ shows). So we can take logarithm either from one or the other but not from both at the same time.