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.
 A: $$
\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c
$$
$$
\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} = g'(x)
$$
$$
\frac 1 {3\sqrt{14}} = g'(2)
$$
$$
\frac 1 {g'(2)} = 3\sqrt{14}, \text{ which is between 11 and 12}
$$
$$
\left\lfloor \frac 1 {g'(2)} \right\rfloor = 11
$$
Addendum :
$$\begin{align}
11=\sqrt{121}&<\sqrt{126}=\sqrt{2\cdot3^2\cdot7}=3\sqrt{14}=\\
&=\sqrt{126}<\sqrt{144}=12
\end{align}$$
$\text{hence ,}$
$$11<3\sqrt{14}<12\;.$$
A: 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 :
$$\begin{align}
11=\sqrt{121}&<\sqrt{126}=\sqrt{2\cdot3^2\cdot7}=3\sqrt{14}=\\
&=\sqrt{126}<\sqrt{144}=12
\end{align}$$
$\text{hence ,}$
$$11<3\sqrt{14}<12\;.$$
