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 $$F'(x) = - \frac{x + 8}{x^3 - 9}$$ ... 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 $$F'(x) = \frac{x + 8}{x^3 - 9}$$ or $$F'(x) = -\frac{x + 8}{x^3 - 9}$$ ??
EDIT: thanks for the answers! but what happens when x is bigger than 9^1/3? The derivative is also the same? Btw, when x = 0 shouldn't F(x) = 0 and hence F'(x) = 0?

We have got the identity $\int\limits_{a}^{b}f(x)dx = -\int\limits_{b}^{a}f(x)dx$. Which can easy be verified by writing it out. Im guessing this is what you mean? 
