Start from:
#arctanx = int_0^x (dt)/(1+t^2)#
Now the integrand function is the sum of a geometric series of ratio #-t^2#:
#1/(1+t^2) = sum_(n=0)^oo (-1)^n(t^2)^n = sum_(n=0)^oo (-1)^nt^(2n)#
so:
#arctan x = int_0^x sum_(n=0)^oo (-1)^nt^(2n)#
This series has radius of convergence #R=1#, so in the interval #x in (-1,1)# we can integrate term by term:
#arctan x = sum_(n=0)^oo int_0^x (-1)^nt^(2n)dt = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)#
and the resulting series has the same radius of convergence.
Now substitute #x= t^3#. the radius of convergence does not change since #abs x < 1 => abs(t^3) < 1#:
#arctan t^3 = sum_(n=0)^oo (-1)^n (t^3)^(2n+1)/(2n+1) = sum_(n=0)^oo (-1)^n t^(6n+3)/(2n+1)#
and integrate again term by term:
#int_0^x arctant^3dt = sum_(n=0)^oo (-1)^n int_0^x t^(6n+3)/(2n+1)dt = sum_(n=0)^oo (-1)^n x^(6n+4)/((6n+4)(2n+1))#
