Split up C into two component paths C₁ and C₂, where each line segment is respectively parameterized by
r₁(t) = (1 - t ) (3i + k) + t (4i + 3j + k) = (t + 3) i + 3t j + k
r₂(t) = (1 - t ) (4i + 3j + k) + t (4i + 5j + 4k) = 4i + (2t + 3) j + (3t + 1) k
both with 0 ≤ t ≤ 1.
It's a bit unclear what function you're supposed to integrate (looks like xyz ?) so I'll give a more general result. The line integral of a scalar function f(x, y, z) along the given path C is
[tex]\displaystyle \int_C f(x,y,z)\,\mathrm ds = \int_{C_1}f(\mathbf r_1(t))\left\|\frac{\mathrm d\mathbf r_1}{\mathrm dt}\right\|\,\mathrm dt + \int_{C_1}f(\mathbf r_2(t))\left\|\frac{\mathrm d\mathbf r_2}{\mathrm dt}\right\|\,\mathrm dt[/tex]
We have
dr₁/dt = i + 3j ==> || dr₁/dt || = √(1² + 3²) = √10
dr₂/dt = 2j + 3k ==> || dr₂/dt || = √(2² + 3²) = √13
Then the integrals reduce to
[tex]\displaystyle \int_0^1 \left(\sqrt{10}\,f(t+3,3t,1) + \sqrt{13}\,f(4,2t+3,3t+1)\right)\,\mathrm dt[/tex]
If indeed f(x, y, z) = xyz, then we have
[tex]\displaystyle \int_0^1 \left(3\sqrt{10}\,t(t+3) + 4\sqrt{13}\,(2t+3)(3t+1)\right)\,\mathrm dt = 11\sqrt{\frac52}+42\sqrt{13}[/tex]
