$$\int \dfrac{\mathrm{arctg}\,x}{1+x^2}\,dx$$

Так как $d(\mathrm{arctg}\,x)=\dfrac{dx}{1+x^2}$: $$\int\mathrm{arctg}\,x\cdot d(\mathrm{arctg}\,x)=\frac{(\mathrm{arctg}\,x)^2}2+C.$$

$$\boxed{\dfrac{(\mathrm{arctg}\,x)^2}2+C}$$