#f(x)=sum_{n=1}^infty(-1)^n{x^{2n+1}}/{2n+1}#
Let us look at some details.
#f(x)=arctanx#
#f'(x)=1/{1+x^2}=1/{1-(-x^2)}#
Remember that the geometric power series
#1/{1-x}=sum_{n=0}^infty x^n#
by replacing #x# by #-x^2#,
#Rightarrow 1/{1-(-x^2)}=sum_{n=0}^infty(-x^2)^n=sum_{n=0}^infty(-1)^nx^{2n}#
So,
#f'(x)=sum_{n=0}^infty(-1)^nx^{2n}#
By integrating,
#f(x)=int sum_{n=0}^infty(-1)^nx^{2n}dx#
by putting the integral sign inside the summation,
#=sum_{n=0}^infty int (-1)^n x^{2n}dx#
by Power Rule,
#=sum_{n=1}^infty(-1)^n{x^{2n+1}}/{2n+1}+C#
Since #f(0)=arctan(0)=0#,
#f(0)=sum_{n=1}^infty(-1)^n{(0)^{2n+1}}/{2n+1}+C=C
Rightarrow C=0#
Hence,
#f(x)=sum_{n=1}^infty(-1)^n{x^{2n+1}}/{2n+1}#
