[tex]\int{\dfrac{1}{2\,{x}^{2}+8}}{\;\mathrm{d}x}=\int{\dfrac{1}{2\,\left({x}^{2}+4\right)}}{\;\mathrm{d}x}\overset{u=x/2}{=}\dfrac{1}{2}\int{\dfrac{2}{4\,{u}^{2}+4}}{\;\mathrm{d}u}=\dfrac{1}{4}\cdot\mathrm{arctg}\left(u\right)+C\\\\\int\limits_{0}^{\infty }{\dfrac{1}{2\,{x}^{2}+8}}{\;\mathrm{d}x}=\lim_{x\to \infty }\frac{1}{4}\mathrm{arctg}\left(\frac{x}{2}\right)-0=\frac{\pi}{8}[/tex]