Compute the differentials:
x = exp(t) ⇒ dx = exp(t) dt
y = 4t ⇒ dy = 4 dt
z = ln(t) ⇒ dz = 1/t dt
Then in the integral, we have
[tex]\displaystyle \int_C ye^z \, dz + x\ln(x) \, dy - y \, dx[/tex]
[tex]\displaystyle = \int_1^2 4te^{\ln(t)} \cdot \frac{dt}t + e^t \ln\left(e^t\right) \cdot 4 \, dt - 4t \cdot e^t \, dt[/tex]
[tex]\displaystyle = \int_1^2 4t \, dt + 4te^t \, dt - 4te^t \, dt[/tex]
[tex]\displaystyle = \int_1^2 4t \, dt = 2(2^2 - 1^2) = \boxed{6}[/tex]
