 Solution

Use $t=\tan u$ with the inverse tangent branch $-\pi /2<u<\pi /2$. Then $dt/(1+t^2)=du$, soFor $|t|<1$, the geometric series gives $(1+t^2{)}^{-1}=1-t^2+t^4-t^6+\cdots$. Integrating term by term on compact subintervals of $(-1,1)$ producesThe integral identity itself holds for every real $x$; the displayed local power-series argument uses $|x|<1$.