[tex]\mathbf G(x,y)=(ye^{xy}+4\cos(4x+y))\,\mathbf i+(xe^{xy}+\cos(4x+y))\,\mathbf j[/tex]
We're computing the line integral
[tex]\displaystyle\int_C\mathbf G\cdot\mathrm d\mathbf r[/tex]
It looks like the circular part of [tex]C[/tex] should be along the circle [tex]x^2+y^2=16[/tex] starting at (4,0) and terminating at [tex]\left(\dfrac4{\sqrt2},\dfrac4{\sqrt2}\right)[/tex].
Because integrating with respect to a parameterization seems like it would be a pain, let's check to see if [tex]\mathbf G[/tex] is a conservative vector field. For this to be the case, if [tex]\mathbf G(x,y)=P(x,y)\,\mathbf i+Q(x,y)\,\mathbf j[/tex], then [tex]\mathbf G[/tex] is conservative iff [tex]\dfrac{\partial P(x,y)}{\partial y}=\dfrac{\partial Q(x,y)}{\partial x}[/tex].
We have [tex]P(x,y)=ye^{xy}+4\cos(4x+y)[/tex] and [tex]Q(x,y)=xe^{xy}+\cos(4x+y)[/tex]. The corresponding partial derivatives are
[tex]\dfrac{\partial P(x,y)}{\partial y}=e^{xy}(1+xy)-4\sin(4x+y)[/tex]
[tex]\dfrac{\partial Q(x,y)}{\partial x}=e^{xy}(1+xy)-4\sin(4x+y)[/tex]
and so the vector field is indeed conservative.
Now, we want to find a function [tex]G(x,y)[/tex] such that [tex]\nabla G(x,y)=\mathbf G(x,y)=\left(\dfrac{\partial G(x,y)}{\partial x},\dfrac{\partial G(x,y)}{\partial y}\right)[/tex]. We have
[tex]\dfrac{\partial G(x,y)}{\partial x}=ye^{xy}+4\cos(4x+y)[/tex]
Integrating with respect to [tex]x[/tex] yields
[tex]\displaystyle\int\frac{\partial G(x,y)}{\partial x}\,\mathrm dx=\int(ye^{xy}+4\cos(4x+y))\,\mathrm dx[/tex]
[tex]G(x,y)=e^{xy}+\sin(4x+y)+g(y)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial G(x,y)}{\partial y}=\dfrac{\partial}{\partial y}\left[e^{xy}+\sin(4x+y)+g(y)\right][/tex]
[tex]xe^{xy}+4\cos(4x+y)=xe^{xy}+\cos(4x+y)+\dfrac{\mathrm dg(y)}{\mathrm dy}[/tex]
[tex]\implies \dfrac{\mathrm dg(y)}{\mathrm dy}=0[/tex]
[tex]\implies g(y)=C[/tex]
and so
[tex]G(x,y)=e^{xy}+\sin(4x+y)+C[/tex]
Because [tex]\mathbf G(x,y)[/tex] is conservative, and a potential function exists, the line integral is path-independent and the fundamental theorem of calculus of line integrals applies, so we can evaluate the line integral by evaluating the potential function at the endpoints. We end up with
[tex]\displaystyle\int_C\mathbf G\cdot\mathrm d\mathbf r=G\left(\frac4{\sqrt2},\frac4{\sqrt2}\right)-G(0,0)=e^8-1+\sin(10\sqrt2)[/tex]
