192. $\int \frac{\operatorname{arctan} x}{1+x^{2}} d x$.
Solution. $\int \frac{\operatorname{arctg} x}{1+x^{2}} d x=\left|\begin{array}{l}\operatorname{arctg} x=t \\ \frac{1}{1+x^{2}} d x=d t\end{array}\right|=\int t d t=\frac{1}{2} t^{2}+C=$ $=\frac{1}{2}(\operatorname{arctg} x)^{2}+C$.