[tex]\displaystyle\int\frac x{\sqrt{16+8x^2}}\,\mathrm dx[/tex]
Let [tex]u=16+8x^2\implies\mathrm du=16x\,\mathrm dx[/tex]. Then the integral becomes
[tex]\displaystyle\frac1{16}\int\frac{\mathrm du}{\sqrt u}\,\mathrm du=\frac18\sqrt u+C[/tex]
[tex]=\dfrac18\sqrt{16+8x^2}+C[/tex]
