[m]\frac{1}{4+x^2}dx=\frac{1}{2}d(arctg\frac{x}{2})[/m]
∫ ^(+ ∞ )_(0)[m](arctg\frac{x}{2})^{-\frac{1}{2}})\cdot \frac{1}{2}d(arctg\frac{x}{2})=[/m]
[m]= \frac{1}{2}\cdot \frac{(arctg\frac{x}{2})^{-\frac{1}{2}+1}}{\frac{1}{2}}[/m]|^(+ ∞ )_(0)=
=sqrt((π/2))-0=sqrt((π/2))