Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$? Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$?

If $\displaystyle I = \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c$, then $\left\lfloor \dfrac{1}{g'(2)} \right\rfloor = \cdots$,
(where $\lfloor \cdot \rfloor$ represents the greatest integer function)

To me it seemed like it's an identity and so it should be safe to differentiate both sides.

Your approach is correct.The formula given clearly indicates that $g(x)$ is the indefinite integral of the function:
$$ \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} $$
So differentiating $g(x)$ gives us back that function of $x$, and evaluating it at $x=2$ after differentiation is correct.
We should get $g'(2) = \frac{1}{3\sqrt{14}}$, and the greatest integer function applied to the reciprocal gives us $\lfloor 3\sqrt{14} \rfloor = 11$.
Addendum :
$$\begin{align}
11=\sqrt{121}&<\sqrt{126}=\sqrt{2\cdot3^2\cdot7}=3\sqrt{14}=\\
&=\sqrt{126}<\sqrt{144}=12
\end{align}$$
$\text{hence ,}$
$$11<3\sqrt{14}<12\;.$$
