212. $\int \arctan x \, dx$.
Solution. $\int \operatorname{arctg} x d x=\left|\begin{array}{l}u=\operatorname{arctg} x, \\ d u=\frac{d x}{1+x^{2}}, v=x\end{array}\right|=x \operatorname{arctg} x-\int \frac{x d x}{1+x^{2}}=$ $=x \operatorname{arctg} x-\frac{1}{2} \int \frac{d\left(1+x^{2}\right)}{1+x^{2}}=x \operatorname{arctg} x-\frac{1}{2} \ln \left(1+x^{2}\right)+C$.