Derivative of Integral help $$\frac{d}{dx}\int_0^x\frac{1}{1+t^4}\,dt$$
Im not sure why but this is confusing me...
Mathematica gives me one answer but I get another...
 A: Hint: Use the Fundamental Theorem of Calculus!
$$F(x) = \int_0^x\!f(t)\, dt \implies \frac {d}{dx}(F(x)) = f(x)$$
Please read the Wikipedia entry (see link above) for some examples as to how to apply it.
This part of the overall theorem is sometimes referred to as the first fundamental theorem of calculus.
The "fine print" (conditions that must be met for the above to hold!):
Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x \in [a, b]$, by
$$F(x) = \int_a^x\!f(t)\, dt.$$
Then, $F$ is continuous on $[a, b]$, differentiable on the open interval $(a, b)$, and
$$F'(x) = f(x)\, \text{ for all } x \in (a, b).$$


With respect to your posted problem, since $f, F$ meet the above conditions, $$\frac{d}{dx}\int_0^x\frac{1}{1+t^4}\,dt = \dfrac{1}{1 + x^4}$$

A: Let $\displaystyle f(x) = \int_{0}^{x}\frac{1}{1+t^4}dt$
Now $\displaystyle \frac{d}{dx}(f(x)) = \frac{d}{dx}\left\{\int_{0}^{x}\frac{1}{1+t^4}dt\right\} = \frac{1}{1+x^4}\cdot \frac{d}{dx}(x)-\frac{1}{1+0^4}\cdot \frac{d}{dx}(0)$
So $\displaystyle f^{'}(x) = \frac{1}{1+x^4}$
