Recall that
[tex]\dfrac{\mathrm d(\arctan x)}{\mathrm dx}=\dfrac1{1+x^2}[/tex]
and that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
So we get
[tex]\displaystyle\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}[/tex]
Taking the integral then gives
[tex]\displaystyle\int\frac{\mathrm dx}{1+x^2}=\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}+C[/tex]
We know that [tex]\arctan0=0[/tex], so if we evaluate both sides of the equation above for [tex]x=0[/tex], we would get a series that converges to 0, so that
[tex]0=0+C\implies C=0[/tex]
and so
[tex]\arctan x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}[/tex]
