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}