You're conveniently given the parameterization of the surface [tex]\mathcal S[/tex],
[tex]\mathbf s(u,v)=\langle u+v,u-v,1+2u+v\rangle[/tex]
where [tex]0\le u\le4[/tex] and [tex]0\le v\le1[/tex], so all we need to set up and compute the integral is to compute the surface element, [tex]\mathrm dS[/tex]:
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm dS=\sqrt{14}\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is
[tex]\displaystyle\iint_{\mathcal S}x+y+z\,\mathrm dS=\sqrt{14}\int_{v=0}^{v=1}\int_{u=0}^{u=4}(u+v)+(u-v)+(1+2u+v)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\sqrt{14}\int_{v=0}^{v=1}\int_{u=0}^{u=4}1+4u+v\,\mathrm du\,\mathrm dv=38\sqrt{14}[/tex]
