To evaluate the integral \(\frac{1}{2\pi j}\oint_C f(z)dz\) where f(z) = \frac{1}{z+1} - \frac{2}{z+3} and C is the path |z+1|=1, we can use the residue theorem from complex analysis.

The residue theorem states that if you have a function f(z) analytic inside and on some simple closed contour C, except for isolated singularities, then:

where the sum is over all residues inside the contour C.

Step-by-step Solution:

1. Identify singularities of f(z):
   • The function \frac{1}{z+1} has a singularity at z = -1.
   • The function \frac{2}{z+3} has a singularity at z = -3.
2. Determine which singularities lie inside the contour |z+1|=1:
   • The circle |z+1|=1 is centered at z = -1 with a radius of 1, so it will include z = -1 but not z = -3.
3. Calculate the residue at z = -1:
   • The function \frac{1}{z+1} has a residue of 1 at z = -1.
   • The function \frac{2}{z+3} does not contribute since z = -3 is not inside the contour.
   • Thus, the total residue inside the contour is simply 1.
4. Apply the residue theorem:
   • The value of the integral is the sum of the residues inside the contour times 2\pi j, thus:
   • \(\frac{1}{2\pi j}\oint_C f(z)dz = 1\).

After evaluating the integral, the correct answer is 1.