As Pringoooals suggested, use Green's Theorem. The key is to integrate $x$ first, then $y$.
$\displaystyle\int_C(y+e^\sqrt{x}) dx + (xe^{y^2}) dy = \iint_{\Omega}\left[\dfrac{\partial}{\partial x}(xe^{y^2}) - \dfrac{\partial}{\partial x}(y+e^{\sqrt{x}})\right]\,dx\,dy = \iint_{\Omega}(e^{y^2}-1)\,dx\,dy = \int_{0}^{2}\int_{0}^{2y}(e^{y^2}-1)\,dx\,dy = \int_{0}^{2}2y(e^{y^2}-1)\,dy = \left[e^{y^2}-y^2\right]_{0}^{2} = e^4-5$.
By the definition of open set, since $(0,0) \in \Omega$, there is an $r>0$ such that the disk of radius $r$ centred at $(0,0)$, $B_r(0,0)$ is contained in $\Omega$. Hence you can define $\Omega_r = \Omega \setminus B_r(0,0)$, which has two distinct boundaries, $\partial \Omega_r$ and $\partial B_r(0,0)$.
Now, as you say, we can use Green's Theorem on $\Omega_r$, so
$$ 0 = \iint_{\Omega_r} (\partial_x Q - \partial_y P)\, dx\,dy = \int_{\partial\Omega_r} P \, dx + Q \, dy = \left( \int_{\partial\Omega} -\int_{x^2+y^2=r^2} \right) (P \, dx + Q \, dy), $$
because the boundary of $B_r(0,0)$ is traversed in the opposite direction because we have to keep the interior of $\Omega_r$ on the left.
Therefore we have to compute $\int_{x^2+y^2=r^2} (P \, dx + Q \, dy)$. Since this is just a circle, we parametrise it by $x=-r\cos{t}$, $y=r\sin{t}$, $dx=r\cos{t} \, dt$, $ dy=r\sin{t} \, dt $:
$$\int_{x^2+y^2=r^2} (P \, dx + Q \, dy) = \int_0^{2\pi} \frac{-r\sin{t}}{r^2(\cos^2{t}+\sin^2{t})} (-r\sin{t}) \, dt + \frac{r\cos{t}}{r^2(\cos^2{t}+\sin^2{t})} (r\cos{t}) \, dt \\
= \int_0^{2\pi} 1 \, dt = 2\pi. $$
Best Answer
Line from $\;(0,0)\to (0,1)\;$ :
$$\ell_1:\;t(0,1)+(1-t)(0,0)=(0,t)\;,\;\;t\in [0,1]\implies$$
$$\int_{\ell_1}Pdx+Qdy=\int_0^1\left((-1)t^2(0\cdot )+(t+1)0^2\right)dt=0$$$${}$$
Line from $\;(0,1)\to \left(\frac12,0\right)\;$ :
$$\ell_2:\;t\left(\frac12,0\right)+(1-t)(0,1)=\left(\frac t2,1-t\right)\;,\;\;t\in [0,1]\implies$$
$$\int_{\ell_2}Pdx+Qdy=\int_0^1\left(\left(\frac t2-1\right)(1-t)^2\cdot\frac12dt+(2-t)\frac{t^2}4(-dt)\right)=$$
$$=\frac14\int_0^1(t-2)(t-1)^2dt+\frac14\int_0^1(t^3-2t^2)dt=\frac14\int_0^1\left(2t^3-6t^2+5t-2\right)dt=$$$${}$$
$$=\frac14\left(\frac12-2+\frac52-2\right)=-\frac14$$
Line from $\;\left(\frac12,0\right)\to (0,0)\;$ :
$$\ell_3:\;t(0,0)+(1-t)\left(\frac12,0\right)=\left(\frac12(1-t),0\right)\;,\;\;t\in [0,1]\implies$$
$$\int_{\ell_3}Pdx+Qdy=\int_0^1\left(-\frac12-t\right)\left(-\frac12\right)+\left(1)0^2\right)dt=\int_0^1\left(\frac14+\frac t2\right)dt=$$
$$=\frac14+\frac14=\frac12$$
so the line integral equals $\;-\cfrac14+\cfrac12=\cfrac14\;$
Using Green's theorem:
$$\frac{\partial Q}{\partial x}=2x(y+1)\;,\;\;\frac{\partial P}{\partial y}=2(x-1)y\implies\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2(x+y)$$
and the integral becomes on the given region:
$$\int_0^{1/2}\int_0^{-2x+1}2(x+y)dydx=\int_0^{1/2}\left(2x(-2x+1)+(2x-1)^2\right)dx=$$
$$=\int_0^{1/2}(-2x+1)dx=-\frac14+\frac12=\frac14$$