Why $\log(-2-y)$ is equivalent to $\log(2+y)$ for restricted values?

Question

I was looking through the step-by-step solution given by Wolfram|Alpha to a problem, and at the last step it says that ...

$= -\log(-2-y(t)) + \mathrm{constant}$

Which is equivalent for restricted t and y values to:

$= -\log(2+y(t)) + \mathrm{constant}$

How come you can equate these two expressions? Or how come you can simply remove the minuses?

Answer 1

The proposed answer is not correct. In fact,

$$\int\frac{y'(t)}{-2-y(t)}dt=-\int\frac{y'(t)}{2+y(t)}dt=-\log|2+y(t)|+K,$$

for all $t$ s.t. $2+y(t)\neq 0$. You can check what does it happen to the r.h.s. of the last equality when you apply the derivative w.r.t. $t$, for both $$2+y(t)> 0,$$

and

$$2+y(t)< 0$$

cases.