studying $\int^x_0\frac{1}{1+t^3}dt$ Let $F(x)=\displaystyle \int^x_0\frac{1}{1+t^3}dt$
1)Prove that F is well defined and differentiable for all $x\in\mathbb{R}$
2)If $n$ is a positive integer show that $F(x)=\displaystyle(\sum\limits_{k=0}^n (-1)^k\frac{x^{3k+1}}{3k+1})+(-1)^{n+1}\int_0^x\frac{t^{3n+3}}{1+t^3}dt$
3)Prove that $F(x)=\displaystyle\sum\limits_{k=0}^\infty (-1)^k\frac{x^{3k+1}}{3k+1}$ for all $|x|\le1$
For part one I'm assuming by well defined it means the integral exists which is direct since the function under integration is continuous, but the differentiability part is not clear I'm not sure how I should approach the limit. As for parts 2,3 I've never encountered any similar problems( I think I should take the limit as $n\to\infty$  in 3) 
 A: Careful about the first statement. The integral makes no sense if $x\leq -1$, since the integrand has an essential singularity at $x=-1$. So you can only work over $x>-1$.
For the third item, you need to determine for which $x$ we have $$R_n(x)=\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\to 0$$
as $n\to \infty$.
Now, note that for $t=-1$ the function is undefined. Since we're expanding throughout the rigin, we can be sure we can't get past that singulatiry, that is, we can be sure that if $x\leq -1$, the series expansion will fail. Now, consider what happens for $-1<t<1$. In that case $t^k\to 0$, so we will exploit this fact to show that the error/remainder does vanish. Note that for $-1<t<1$ (so $-1<x<1$), $1+t^3>0$, so that
$$0\leq \left|\int_0^x \frac{t^{3n+3}}{1+t ^3}dt\right|\leq \int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt$$
Now, if we fix an $-1<x<1$, the sequence of functions 
$$f_n(t)=\frac{|t|^{3n+3}}{1+t ^3}$$
converges uniformly to $0$ over $[-x,x]$, in the sense that given any $\epsilon >0$, we can take $n$ large enough so that no matter what $t\in[-x,x]$ we take 
$$\frac{|t|^{3n+3}}{1+t ^3}<\epsilon $$ 
This is merely because $|t|^{3n+3}\to 0$ and because the function is increasing in both directions of the axis. This only means that for any $-1<x<1$, and taking $n$ large enough, we will have $$\int_0^x \frac{|t|^{3n+3}}{1+t ^3}dt<x\epsilon $$ so that $R_n\to 0$, as desired. Now, try to show that for $x\geq 1$, the error term does not go to zero, so that the power series doesn't converge to $f$.
A: For 1) use the Fundumental Theorem of Calculus
For 2) and 3) use that
$$\frac{1}{1+t^3}=\sum\limits_{k=0}^{n}(-1)^k t^{3k}+\frac{(-1)^{n+1}t^{3n+3}}{1+t^3}$$
after proving it (easy proof if you expand the summations).
EDIT: More detail:
1) Because $f(t)=\frac{1}{1+t^3}$ is continuous in $[0,x]$ by FCT, $F$ is differentiable in $[0,+\infty)$.
2) Taylor's Theorem.
