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?

Why do you presume we can't evaluate an integral going from a smaller number to a bigger number? The restriction on being positive is for areas, not integrals. The upper and lower limits can be any numbers, within the specified set of numbers(real or complex).
And no, $F'(x)$ has the same expression for all conditions, of course it may take different values.
