Find the sum using term-by-term differentiation:
I just don't know how to deal with the term. Taking the third derivative I get, for
It is not hard to see that
.
Here we have for all .
For this power series, we see that the radius of convergence is .
Hence converges for all , you may check the end points.
Thus, this power series converges uniformly in for any , hence term-by-term differentiable (and integrable) on the same interval.
As we may pick arbitrarily closer to , we get
........
........
........
for all .
See Uniform Convergence -- from Wolfram MathWorld