= Solution
Use $t=\tan u$ with the inverse tangent branch $-\pi/2<u<\pi/2$. Then $dt/(1+t^2)=du$, so
$$
\boxed{\int_0^x\frac{dt}{1+t^2}=\arctan x.}
$$
For $|t|<1$, the <geometric series> gives $(1+t^2)^{-1}=1-t^2+t^4-t^6+\cdots$. Integrating term by term on compact subintervals of $(-1,1)$ produces
$$
\boxed{\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+O(x^9).}
$$
The integral identity itself holds for every real $x$; the displayed local power-series argument uses $|x|<1$.
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