Derivative of an integral from 0 to x when x is negative?

Given a function $F(x)={\int}_{0}^{x}\frac{t+8}{{t}^{3}-9}dt,$, is F′(x) different when $x<0$, when $x=0$ and when $x>0$?

When $x<0$, is ${F}^{\prime}(x)=-\frac{x+8}{{x}^{3}-9}$.

... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.

EDIT: Yes, I was talking about that identity. So is ${F}^{\prime}(x)=\frac{x+8}{{x}^{3}-9}$ or ${F}^{\prime}(x)=-\frac{x+8}{{x}^{3}-9}$.

Given a function $F(x)={\int}_{0}^{x}\frac{t+8}{{t}^{3}-9}dt,$, is F′(x) different when $x<0$, when $x=0$ and when $x>0$?

When $x<0$, is ${F}^{\prime}(x)=-\frac{x+8}{{x}^{3}-9}$.

... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.

EDIT: Yes, I was talking about that identity. So is ${F}^{\prime}(x)=\frac{x+8}{{x}^{3}-9}$ or ${F}^{\prime}(x)=-\frac{x+8}{{x}^{3}-9}$.