Taylor series expansion is not the best way to approach this problem as it gives an infinite series even for small values of n. Its just an integration by parts problem.
I=[1/(n+1)]x^(n+1)arctanx-[1/(n+1)]∫x^(n+1)/(1+x^2)dx
A reduction can be formed for ∫x^(n+1)/(1+x^2)dx:
J[n+1]=∫x^(n+1)/(1+x^2)dx =∫x^2*x^(n-1)/(1+x^2)dx
=∫[x^(n-1)-[x^(n-1)/(1+x^2)]]dx=(1/n)x^n-J[n-1]
the end result is rather messy for large n.