Change both the variable and the limits

Square first: $u^2=x-2$, so $x=u^2+2$ and $2u\,du/dx=1$, equivalently $dx/du=2u$. At $x=4$, $u=\sqrt2$; at $x=8$, $u=\sqrt6$. Thus the integral becomes $\int_{\sqrt2}^{\sqrt6}\frac{u^2+3}{(u^2+2)u}\,2u\,du$.

Cancel $u$ and divide: $2\int_{\sqrt2}^{\sqrt6}\frac{u^2+3}{u^2+2}\,du=2\int_{\sqrt2}^{\sqrt6}(1+\frac1{u^2+2})\,du=2[u+\frac1{\sqrt2}\tan^{-1}(\frac u{\sqrt2})]_{\sqrt2}^{\sqrt6}$.

The inverse tangent values are principal values in radians: $\tan^{-1}\sqrt3=\pi/3$, $\tan^{-1}1=\pi/4$. The result is $2\sqrt6+\sqrt2(\pi/3)-2\sqrt2-\sqrt2(\pi/4)=2\sqrt6-2\sqrt2+\frac{\pi\sqrt2}{12}$.