4. $\operatorname{arctg} x+\operatorname{arctg} \frac{1-x}{1+x}$ 的值是

4. The value of $\operatorname{arctg} x+\operatorname{arctg} \frac{1-x}{1+x}$ is
4. $-\frac{3}{4} \pi$ or $\frac{\pi}{4}$.

Let $\operatorname{arctg} x=\alpha, \operatorname{arctg} \frac{1-x}{1+x}=\beta$. Then
$$
\begin{array}{l}
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{x+\frac{1-x}{1+x}}{1-x \cdot \frac{1-x}{1+x}} \\
=\frac{x+x^{2}+1-x}{1+x-x+x^{2}}=1 .
\end{array}
$$

When $x>-1$, $\frac{1-x}{1+x}>-1$, then $\alpha, \beta \in\left(-\frac{\pi}{4}\right.$, $\left.\frac{\pi}{2}\right)$, $\alpha+\beta \in\left(-\frac{\pi}{2}, \pi\right)$. Hence $\alpha+\beta=\frac{\pi}{4}$.

When $x<-1$, $\frac{1-x}{1+x}<-1$, then $\alpha, \beta \in\left(-\frac{\pi}{2}\right.$, $\left.-\frac{\pi}{4}\right)$, $\alpha+\beta \in\left(-\pi,-\frac{\pi}{2}\right)$. Hence $\alpha+\beta=-\frac{3}{4} \pi$.