Answer :
[tex]\displaystyle\int_{\mathcal C}z\,\mathrm dx+x\,\mathrm dy+y\,\mathrm dz[/tex]
[tex]x=t^3\implies\mathrm dx=3t^2\,\mathrm dt[/tex]
[tex]y=t^4\implies\mathrm dy=4t^3\,\mathrm dt[/tex]
[tex]z=t^3\implies\mathrm dz=3t^2\,\mathrm dt[/tex]
So the integral is equivalent to
[tex]\displaystyle\int_{t=0}^{t=1}(3t^2+4t^3+3t^2)\,\mathrm dt=3[/tex]
[tex]x=t^3\implies\mathrm dx=3t^2\,\mathrm dt[/tex]
[tex]y=t^4\implies\mathrm dy=4t^3\,\mathrm dt[/tex]
[tex]z=t^3\implies\mathrm dz=3t^2\,\mathrm dt[/tex]
So the integral is equivalent to
[tex]\displaystyle\int_{t=0}^{t=1}(3t^2+4t^3+3t^2)\,\mathrm dt=3[/tex]