It looks like the integral is
[tex]\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy[/tex]
where C is the circle of radius 2 centered at the origin.
You can compute the line integral directly by parameterizing C. Let x = 2 cos(t ) and y = 2 sin(t ), with 0 ≤ t ≤ 2π. Then
[tex]\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \int_0^{2\pi} \left((3x(t)+4y(t))\dfrac{\mathrm dx}{\mathrm dt} + (2x(t)-3y(t))\frac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt \\\\ = \int_0^{2\pi} \big((6\cos(t)+8\sin(t))(-2\sin(t)) + (4\cos(t)-6\sin(t))(2\cos(t))\big)\,\mathrm dt \\\\ = \int_0^{2\pi} (12\cos^2(t)-12\sin^2(t)-24\cos(t)\sin(t)-4)\,\mathrm dt \\\\ = 4 \int_0^{2\pi} (3\cos(2t)-3\sin(2t)-1)\,\mathrm dt = \boxed{-8\pi}[/tex]
Another way to do this is by applying Green's theorem. The integrand doesn't have any singularities on C nor in the region bounded by C, so
[tex]\displaystyle \int_C (3x+4y)\,\mathrm dx + (2x-3y)\,\mathrm dy = \iint_D\frac{\partial(2x-3y)}{\partial x}-\frac{\partial(3x+4y)}{\partial y}\,\mathrm dx\,\mathrm dy = -2\iint_D\mathrm dx\,\mathrm dy[/tex]
where D is the interior of C, i.e. the disk with radius 2 centered at the origin. But this integral is simply -2 times the area of the disk, so we get the same result: [tex]-2\times \pi\times2^2 = -8\pi[/tex].
