1465. Expand $\operatorname{arctg} x$ into a Maclaurin series.
Solution. It is known that $\operatorname{arctg} x=\int_{0}^{x} \frac{d x}{1+x^{2}}$.

We expand the integrand into a series using the binomial series (4) § 4:

$$
\frac{1}{1+x^{2}}=\left(1+x^{2}\right)^{-1}=1-x^{2}+x^{4}-x^{6}+\ldots
$$

Integrating this series within its interval of convergence $(-1,1)$, we obtain

$$
\begin{aligned}
& \operatorname{arctg} x=\int_{0}^{x}\left(1-x^{2}+x^{4}-x^{6}+\ldots\right) d x= \\
& =x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\ldots(-1 \leqslant x \leqslant 1)
\end{aligned}
$$