Fundamental theorem of Calculus: $\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt$ Possible textbook mistake using the FTC we are supposed to evaluate
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
$$
and the answer, the textbook says, should be a constant. When I evaluate that derivative I get
$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt = \frac{d}{dx} \int_{-x}^{0} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{d}{dx} \int_{0}^{-x} \frac{1}{3+t^2} \ dt + \frac{d}{dx} \int_{0}^{x} \frac{1}{3+t^2} \ dt\\
= - \frac{1}{3+(-x)^2} \cdot (-1) + \frac{1}{3+x^2}\\
= \frac{1}{3+x^2} + \frac{1}{3+x^2}\\
= \frac{2}{3+x^2}
$$
but I suppose that the textbook is expecting the answer $0$ (which for me is not correct since the chain rule, on the first term, makes the fraction not cancel). Is there really a mistake?
Thank you.

$$
\frac{d}{dx} \int_{-x}^{x} \frac{1}{3+t^2} \ dt
= 1\frac{1}{3+x^2}+1\frac{1}{3+(-x)^2}=\frac{2}{3+x^2}$$
Note that $$
\frac{d}{dx} \int_{g(x)}^{f(x)} h(t) \ dt=f'(x)h(f(x))-g'(x)h(g(x))$$
